
Glass. 



/CiA ■=,*■* 



Book.: 



Ho. 



AN 



ELEMENTARY TREATISE 



DIFFERENTIAL AND INTEGRAL 
CALCULUS, 

WITH EXAMPLES AND APPLICATIONS. 

BY 

GEORGE A. OSBORNE, S.B., 

Pbofessor of Mathematics in the Massachusetts 
Institute of Technology. 



««< 



D. C. HEATH & CO., PUBLISHERS 

BOSTON NEW YORK CHICAGO 
1903 



i4 



COPYBIGHT, 1891, 

Bt GEORGE A. OSBORNE 



466555 



PEEFACE. 



This work, intended as a text-book for colleges and scien- 
tific schools, is based on the method of limits, as the most 
rigorous and most intelligible form of presenting the first 
principles of the subject. The method of limits has also the 
important advantage of being a familiar method; for it is 
now so generally introduced in the study of the more ele- 
mentary branches of mathematics, that the student may be 
assumed to be fully conversant with it on beginning the 
Differential Calculus. 

The rules or formulae for differentiation in Chapter III. 
differ in one respect from those in similar text-books, in being 
expressed in terms of u instead of x, u being any function 
of x. They are thus directly applicable to all expressions, 
without the aid of the usual theorem concerning a function of 
a function. 

After acquiring the processes of differentiation, the student 
in Chapter V. is introduced to the differential notation, as a 
convenient abbreviation of the corresponding expressions by 
differential coefficients. This notation has manifest advan- 
tages in the study of the Integral Calculus and in its 
applications. 

In Chapter IX. and subsequent pages I have introduced for 

Partial Differentiation the notation — , which has recently 

come into such general use. 

iii 



iv PREFACE. 

The chapters on Maxima and Minima have been placed 
after the applications to curves, as the consideration of that 
subject is much simplified by representing the function by 
the ordinate of a curve. Maxima and Minima may be taken, 
if desired, with equal advantage immediately after Chapter 
XIII. 

In Chapter X., Integral Calculus, I have taken the problem 
of finding the Moment of Inertia of a plane area, as a better 
illustration of double integration than that of finding the 
area itself. The student more readily comprehends the inde- 
pendent variation of x and y in the double integral, 

I I 0& 2 + y 2 )dxdy, than in J I dxdy. 

A few pages of Chapter XII., Integral Calculus, are devoted 
to a description of the Hyperbolic Functions together with 
their differentials, and a comparison is made with the cor- 
responding Circular Functions. 

G. A. OSBORNE. 
Boston, 1895. 



CONTENTS. 



DIFFERENTIAL CALCULUS. 

Chapter I. 

Abts. FUNCTIONS. PJGE9 . 

1-4. Definition and Classification of Functions 1,2 

5. Notation of Functions. Examples 3, 4 

Chapter II. 

DIFFERENTIAL COEFFICIENT. 

6, 7. Limit. Increment . . . . . 5 

8-10. Differential Coefficient. Examples 6-9 

Chapter III. 

DIFFERENTIATION. 

11-13. Differentiation of Algebraic Functions. Examples 10-21 

14-16. Differentiation of Logarithmic and Exponential Functions. 

Examples 21-27 

17,18. Differentiation of Trigonometric Functions. Examples.. 27-32 
19, 20. Differentiation of Inverse Trigonometric Functions. Ex- 
amples 32-37 

21, 22. Differentiation of Inverse Function and Function of a 

Function. Examples 37-40 

Chapter IV. 
SUCCESSIVE DIFFERENTIATION. 

23, 24. Definition and Notation 41 

25. The nth Differential Coefficient. Examples 42-45 

26. Leibnitz's Theorem. Examples 45-47 

v 



VI CONTENTS. 



Chapter V. 

Abts. DIFFERENTIALS. PjGE8 

27. Differentials as related to Differential Coefficients 48, 49 

28. Differentiation by Differentials 49 

29. Successive Differentials. Examples 50, 51 

Chapter VI. 
IMPLICIT FUNCTIONS. 

30. Differentiation of Implicit Functions. Examples 52-54 

Chapter VII. 

EXPANSION OE EUNCTIONS. 

32-36. Maclaurin's Theorem. Examples 55-60 

37-41. Taylor's Theorem. Examples 60-63 

42-45. Eigorous Proof of Taylor's Theorem 64, 65 

46-49. Remainder in Taylor's and Maclaurin's Theorems 66-68 

Chapter VIII. 
INDETERMINATE FORMS. 

50, 51. Limiting Value of a Fraction 69 

52, 53. Evaluation of x« Examples . 70-73 

54-57. Evaluation of §, oo, oo — go. Examples 73-76 

58. Evaluation of Exponential Forms. Examples 76-78 

Chapter IX. 
PARTIAL DIFFERENTIATION. 

59, 60. Partial Differential Coefficients of First Order. Exam- 
ples 79, 80 

61-63. Partial Differential Coefficients of Higher Orders. Exam- 
ples 80-82 

64,65. Total Differential of Functions of Several Variables. 

Examples 82-84 

66. Condition for an Exact Differential. Examples 85 

67. Differentiation of Implicit Functions 86 

68, 69. Taylor's Theorem for Several Variables 87, 88 



CONTENTS. VL1 



Chapter X. 

CHANGE OF VARIABLES IN DIFFERENTIAL 

COEFFICIENTS. 

Arts. Pages. 

70. Changing from x to y 89 

71, 72. Changing from y to z 90 

73. Changing from x to z. Examples 90-92 



Chapter XI. 

REPRESENTATION OF VARIOUS CURVES. 

74-85. Rectangular Co-ordinates 93-98 

86-93. Polar Co-ordinates 98-102 

Chapter XII. 

DIRECTION OF CURVE. TANGENT AND NORMAL. 
ASYMPTOTES. 

94-97. Direction of Curve. Subtangent and Subnormal. 

Examples 103-108 

98, 98£. Differential Coefficient of the Arc 108,109 

99. Equation of the Tangent and Normal. Examples . . . 109-112 

100-106. Asymptotes. Examples 112-116 

Chapter XIII. 

DIRECTION OF CURVATURE. POINTS OF INFLEXION. 

107-109. Direction of Curvature 117 

110. Points of Inflexion. Examples „ . . 118, 119 

Chapter XIV. 

CURVATURE. CIRCLE OF CURVATURE. EVOLUTE 
AND INVOLUTE. 

111-113. Definition of Curvature ; Uniform and Variable 120, 121 

114, 115. Radius of Curvature. Examples 121-124 

116. Centre of Curvature 124, 125 

117-121. Evolute and Involute. Examples 125-128 



Vlll CONTENTS. 



Chapter XV. 



ORDER OF CONTACT. OSCULATING CIRCLE. 

Aets. Pages. 

122, 123. Consecutive Common Points 129, 130 

124, 125. Osculating Curves , . 130, 131 

126-128. Analytical Conditions for Contact 131-133 

129, 130. Osculating Circle. Examples 133-136 

Chapter XVI. 

ENVELOPES. 

131-133. Series of Curves. Definition of Envelope 137, 138 

134-136. Equation of Envelope 138-140 

137. Evolute, the Envelope of Normals. Examples 140-144 

Chapter XVII. 

SINGULAR POINTS OF CURVES. 

138-141. Multiple Points 145-148 

142, 143. Points of Osculation. Cusps 149, 150 

144. Conjugate Points. Examples 150-152 

Chapter XVIII. 

MAXIMA AND MINIMA OF FUNCTIONS OF ONE 
INDEPENDENT VARIABLE. 

145-149. Definition. Conditions for Maxima and Minima de- 
rived from Curves 153-157 

150, 151. Conditions for Maxima and Minima by Taylor's 

Theorem. Examples 157-162 

Problems in Maxima and Minima 162-164 

Chapter XIX. 

MAXIMA AND MINIMA OF FUNCTIONS OF SEVERAL 
INDEPENDENT VARIABLES. 

152-155. Definition. Conditions for Maxima and Minima by 

Taylor's Theorem. Examples 165-171 



CONTENTS. ix 

1 

INTEGRAL CALCULUS. 
Chapter I. 

Arts> elementary forms of integration. Pages 

1, 2. Definition of Integration. Elementary Principles 173, 174 

3. Fundamental Integrals 175, 176 

4-7. Derivation and Application of Fundamental Formulae. 

Examples 176-187 

Chapter II. 
INTEGRATION OF RATIONAL FRACTIONS. 

8, 9. Preliminary Operation. Factors of Denominator 188, 189 

10. Case I. Examples 189-191 

11. Case II. Examples 191,192 

12. Case III. Examples 192-195 

13. Case IV. Examples 195-198 

Chapter III. 

INTEGRATION BY RATIONALIZATION. 

14-16. Fractional Powers of x and of a + bx. Examples 199-201 

17. Fractional Powers of a + bx 2 . Examples 201, 202 

18,19. Expressions containing V± x' 1 + ax + b. Examples... 202-204 

20. Integration by Substitution. Examples 204, 205 

Chapter IV. 

INTEGRATION BY PARTS. INTEGRATION BY SUCCESSIVE 
REDUCTION. 

21. Integration by Parts. Examples 206-208 

22-24. Formulae of Reduction. Examples 208-214 

Chapter V. 

TRIGONOMETRIC INTEGRALS. 

25-27. Integration of tan n xdx; of sec n xdx; of tan^x sec":*; dx. 

Examples 215-218 

28, 29. Integration of sm m x cos H a; dx. Examples 219-222 



X CONTENTS. 

Arts. Pages. 
30. Trigonometric, transformed into Algebraic, Integrals. 

Examples 222-224 

31, 32. Trigonometric ForniulEe of Reduction. Examples 224-226 

33-35. Integration of — — and — ; 

a+b sin x a+b cos x 

of e ax sin nxdx and e ax cos nxdx. 
Examples 226-229 

Chapter VI. 
INTEGRALS FOR REFERENCE. 



36. Integrals containing Va' 2 — x 2 ; Vx' 2 ±a' 2 ; ± ax 2 + bx + c. 230-235 

Chapter VII. 

INTEGRATION AS A SUMMATION. DEFINITE INTEGRALS. 

37-40. Integration, the Summation of an Infinite Series 236-240 

41-43. Definition of Definite Integral. Examples 240-244 

Chapter VIII. 

APPLICATION OF INTEGRATION TO PLANE CURVES. 

APPLICATION TO CERTAIN VOLUMES. 

44-47. Areas of Curves. Examples 245-249 

48, 49. Lengths of Curves. Examples 249-252 

50, 51. Surfaces of Revolution. Examples 252-255 

52. Other Volumes. Examples 255-257 

Chapter IX. 

SUCCESSIVE INTEGRATION. 

53-56. Double and Triple Integrals. Examples 258-260 

Chapter X. 

DOUBLE INTEGRATION APPLIED TO PLANE AREAS 
AND MOMENT OF INERTIA. 

57-60. Double Integration. Rectangular Co-ordinates. Ex- 
amples 261-264 

61-63. Double Integration. Polar Co-ordinates. Examples . . 264-266 



CONTENTS. XI 

Chapter XI. 

SURFACE AND VOLUME OF ANY SOLID. __ a 

Arts. Pages. 

64, 65. Area of any Surface. Examples 267-270 

66, 67. Volume of any Solid. Examples 270-273 

Chapter XII. 

HYPERBOLIC FUNCTIONS. CYCLOID, EPICYCLOID, AND 
HYPOCYCLOID. INTRINSIC EQUATION OF A CURVE. 

69-71. Definitions of Hyperbolic, and Inverse Hyperbolic, 

Functions 274-276 

72, 73. Differentiation of Hyperbolic Functions. Inverse Hyper- 
bolic Functions as Integrals 276, 277 

74, 75. Hyperbolic Functions and the Hyperbola. Exercises . . 278-280 

76-82. Equatiop and Properties of the Cycloid 280-284 

83-89. Equations and Properties of the Epicycloid, and Hypo- 
cycloid 284-288 

90-93. Intrinsic Equation of a Curve and of its Evolute. 

Examples 289-292 



DIFFERENTIAL CALCULUS. 

CHAPTER I. 

FUNCTIONS. 

1. Definition of a Function. When the value of one variable 
quantity -%& depends upon that of another, jhat any change 
in the latter produces a corresponding change in the former, 
the former is said to be a function of the latter. 

For example, the area of a square is a function of its side ; 
the volume of a sphere is a function of its radius ; the sine, 
cosine, and tangent are functions of the angle j the expressions 



x 2 , log(^ + l), V»(aj + 1), 
are functions of x. 

A quantity may be a function of two or more variables. 
For example, the area of a rectangle is a function of two 
adjacent sides ; either side of a right triangle is a function of 
the two other sides ; the volume of a rectangular parallelo- 
piped is a function of its three dimensions. 

The expressions 

x 2 + xy + y 2 , log (x 2 + y 2 ) , a* + «, 

are functions of x and y. 
The expressions 

xy + yz+zx, \p+^, log (x 2 + y- »), 

i * z 

are functions of x, y, and z. 

2. Dependent and Independent Variables. If y is a function 
of x, as in the equations 

y = x 2 , y = tan Ax, y = e x , 



2 DIFFERENTIAL CALCULUS. 

x is called the independent variable, and y the dependent 
variable. 

It is evident that whenever y is a function of x, x may be 
also regarded as a function of y, and the positions of dependent 
and independent variables reversed. Thus from the preceding 
equations, _ 

x = Vy, x = i tan- 1 y, x = log e y. 

In equations involving more than two variables, as 

z-\-x — y = 0, w + wz + zx + y = 0, 

one must be regarded as the dependent variable, and the 
others as independent variables. 

3. Explicit and Implicit Functions. When one quantity is 
expressed directly in terms of another, the former is said to 
be an explicit function of the latter. 

For example, y is an explicit function of x in the equations, 



y = x 2 -\-2x, y = V# 2 + 1. 

When the relation between y and x is given by an equation 
containing these quantities, but not solved with reference to y, 
y is said to be an implicit function of x, as in the equations, 

2xy + y 2 = x 2 + 1, y + logy = x. 

Sometimes, as in the first of these equations, we can solve 
the equation with reference to y, and thus change the function 
from implicit to explicit. Thus we find from this equation, 



y = -x±V2x 2 + l. 

4. Algebraic and Transcendental Functions. An algebraic 
function is one that involves only the operations of addition, 
subtraction, multiplication, division, involution and evolution 
with constant exponents. All other functions are called tran- 
scendental functions, including logarithmic, exponential, trigo- 
nometric, and inverse trigonometric, functions. 



FUNCTIONS. 3 

5. Notation of Functions. The symbols F(x), f(x), <f>(x), 
i}/(x), and the like, are used to denote functions of x. Thus 
instead of " y is a function of x" we may write 

y=f(x) or y = <j>(x). 

A functional symbol occurring more than once in the same 
problem or discussion is understood to denote the same func- 
tion or operation, although applied to different quantities. 
Thus, if 

f(x)=x> + S, (1) 

then f(y) = y 2 + 5, f(a) = a 2 + 5, 

f(a + 1) = (a + 1) 2 + 5 = a 2 + 2 a + 6, 
/(2) = 2 2 +'5 = 9, /(1) = 6. 

In all these expressions /( ) denotes the same operation as 
denned by (1) ; that is, the operation of squaring the quantity 
and adding 5 to the result. 

The following examples will further illustrate the notation 
of functions. 

EXAMPLES. 

1. If f(x) = 2x* — x 2 -7x + 6, show that 
A3) = 30, /(2) = 4, /(0) = 6, /(1) = 0, 
/(-2) = 0, /(f) = 0, f(x-2) = 2x*-13x 2 + 21x, 
f(x + h) = 2x i + (6h-l)x 2 + (6h 2 -2h-7)x + 2h s 

-.h 2 -7h + 6. 

2. Given f 1 (y) = 2y i -y s +l, f 2 (y) = 7y*-6y+V, show that 

/i(l)=/*(l), /i(|)=/.(l), /i(-2)=/ 2 (-2), 

/i(0)=/i(0). 

3. If /(a) = ~ , show that 

Qj -J- J- 



f(a)-f(b) = a-b 
l+f(a)f(fi) 1 + ab 



4 DIFFERENTIAL CALCULUS. 

4. If <j>(m) = (m + 1) m(m — 1) (m — 2), show that 
<j>(2) = <f>(l) = <f>(0) = <f>(-l) = 0, <f>(3) = 4>(-2), 

<f>(m + 1) _ <H m ) 
m + 2 m — 2 

5. If <£(#) = (a? — a) (x — b) (x — c), show that 
*(<*)= *(&)=-+(c) = 0, 

W(0)] s 9 V 2 / 

^>(-a).^(-5).^(-c) =8 )]2< 

<£(0) L V yj 

6. If <j>(u) = e u + e~ u , show. that 

*(3tf). = [>(*)] 8 -3*(i0, 

$(u + v)$(u — v)=<j>(2u) + cj>(2v), 

7. If F(x) = log 1 ^^, show that 

1 -f- x 

F(x) + F(z) = f(^±^\ 

8. If /(a?) = log (x + Var 9 -1), show that 

2f(x)=f(2x*-l), 
3f(x)=f(4.x*-3x). 

9. Given ^(a;) = cossc + V — 1 sina;; show that 

$ (2 a) = [>(a)] 2 , *(a + 5) = ^(a) ^r(&). 

10. If f(x, y, z) = or 3 -f j/ 3 + z 3 — 3 <eyz, show that 

where L =± px -J- qy + rz, 

3I=py + qz + ra;, 
N = pz +qx-\- ry. 



CHAPTER II. 

DIFFERENTIAL COEFFICIENT. 

6. Limit. The limit of a variable quantity is a fixed value 
or condition, -from which it can be made to differ as little as 
we please. ^aaA\ ^^^X^'VvsA/ 

The student is supposed to be already familiar with the 
meaning of this term, of which the following illustrations 
may be mentioned. 

The limit of the value of the recurring decimal .3333 •••,. 
as the number of decimal places is indefinitely increased, is \. 

The limit of the sum of the series 1 + \ -f- £+ \ -\ , as 

the number of terms is indefinitely increased, is 2. 

The circle is the limit of a regular polygon, as the number 
of sides is indefinitely increased. 

The tangent to a curve is the limit of a secant, as the points 
of intersection approach coincidence. 

The limit of the fraction, m , as approaches zero, is 1, — 

6 

provided 6 is expressed in circular measure. 

7. Increments. An increment of a variable quantity is any 
addition to its value, and is denoted by the symbol A written 
before this quantity. Thus Ax denotes an increment of x, 
Ay an increment of y. 

For example, if we have given 

y = x 2 , 

and assume x = 10, then if we increase the value of x by 2, 
the value of y is increased from 100 to 144, that is, by 44. 

In other words, if we assume the increment of x to be Ax =2, 
we shall find the increment of y to be Ay = 44. 



DIFFERENTIAL CALCULUS. 



A negative increment is a decrement; that is, a decrease in 
valne. 

For example, calling x = 10, as before, in y = x 2 . 



if Aa = — 2, 



then Ay = — 36. 



8. Differential Coefficient. In the equation y = x*, if we 
suppose a? to vary, y will vary also. To fix the attention upon 
a definite value of x, let us suppose x = 10 and therefore 
2/ = 100, and let us inquire what addition or increment will 
be produced in y by a certain increment assigned to x. Cal- 
culating the values of Ay corresponding to different values of 
Aa?, we find results as in the following table : 



If Aa? = 


then Ay = 


and ~ = 
Ax 


3. 


69. 


23. 


2. 


44. 


22. 


1. 


21. 


21. 


0.1 


2.01 


20.1 


0.01 


0.2001 


20.01 


0.001 


0.020001 


20.001 


h. ' 


20& + h\ 


20 + h. 



The third column gives the value of the ratio between the 
increments of x and of y. 

It appears from the table that, as Ax diminishes and 
approaches zero, Ay also diminishes and approaches zero. 

Aw 

The ratio — — diminishes, but instead of approaching zero, ap- 
i\x 

proaches 20 as its limit. 

Ay 
This limit of -r- is called the differential coefficient of y with 



DIFFERENTIAL COEFFICIENT. 7 

respect to x, and is denoted by _^. In this case, when x = 10, 

, dx 

^ = 20. 
ax 

di/ 
It is to be noticed that -=- is not here defined as a fraction, 
cix 

bnt as a single symbol denoting the limit of the fraction 

~ • The student will find as he advances that -^- has many 

Ax dx J 

of the properties of an ordinary fraction, and Chapter V. shows 
how it may be regarded as such. 

9. Without restricting ourselves to any one numerical value, 

dy 
we may obtain -~ from the equation y = x 2 thus : 

ClX 

Having y = x 2 , let Ax = h, and let the new value of y be 
denoted by 

y.'= (« + *)•; 
therefore 

Ay = y'-y = (x + Jl) 2 - X 2 =2xh + h 2 . 

Dividing by Ax = h, gives 

Ax 

The limit of this, when h approaches zero, is 2x. Hence 

dx 

In the same way the differential coefficients of other given 

functions may be found. 

dy 
For example, find ~ from the equation, 

CLX 

y = 2x 3 +l. 
Let Ace = h, 

then y'=2(x + hy+l. 

Ay = y'- y = 2(x + h) 3 - 2a 3 = 2(3 x 2 h + 3xh'+ ft 3 ). 



DIFFERENTIAL CALCULUS. 



Dividing by Ax == h gives 

^ = 2(3x 2 +3xh + h 2 ). 

The limit of -^ is 6 a? 2 , as h approaches zero. 





ax 




Take for another example 






y = ^/x. 


Aa; = 7&. 




y'=z ViC + h. 






Ay = Va; -f- /* — Va?. 






Ay _ V# 4- ^ — Vx 






Ax h 




The limit of this takes the indeterminate form 


rationalizing the numerator, we have 




&y _ 


h 


1 


Ax 


^(Vx + /i+Vx) Vx 


+ h + -Vx 


■ The limit of 


A?/_ 1 . 

&x 2Vx' 




that is, 


dx 2V^ 





But by 



10. General Definition of Differential Coefficient. 
In general, if y = <j>(x), 

y'=<J>(x + 7i), 

&y = y'-y = <t>(x + h)- <f>(x), 

Ay_ <f>(x + h) — <l>(x) ^ 
Ax h 

-^ = limit of ^ — — — ' ^ ' j as h approaches zero. 
dx h ri 



The differential coefficient of a function may then be denned 



DIFFEBENTIAL COEFFICIENT. 9 

as the limiting value of the ratio of the increment of the function 
to the increment of the variable, as these increments approach 
zero. That is, the differential coefficient of the function <f>(x) 
with respect to x, is ^ / \y~ cyH* ,_#^ 

the limit of *(* + *)-* !*),' 
h 

as h is indefinitely diminished. 

The differential coefficient is sometimes called the derivative. 

Note. — In Art. 94 will be found a geometrical illustration 
of the differential coefficient. 

EXAMPLES. 

Following the process of Art. 9, derive the following dif- 
ferential coefficients : 



. 1. y = 3x?-2x. 


dx 


2. y = x*+5. 


^ = 4*1 

dx 


3. y = (x-l)(2x + 3). 


dx 


A 1 
4. y = — 

u X 


dx x 2 ' 


5. y = ¥ 


dy 2 a 
dx~ x 3 " 


x — a 
6. y = 


dy 2 a 
dx~ (x + a) 2 * 


7. y = x%. 


dy 3x* 
dx~~2~' 




dy x 


8. y = -y/x*-2. 




dx - y /x 1 -2 


9 y- 2 • 


dy 1 


Vz + l 

10. y = x J . 


dx (a; + 1)1" 

dx 3 J 



CHAPTER III. 

DIFFERENTIATION. 

11. The process of finding the differential coefficient of a 
given function is called differentiation. The examples in the 
preceding chapter are introduced to illustrate the meaning 
of the differential coefficient, but this elementary method of 
differentiation is too tedious for general use. 

Differentiation is more readily performed by the application 
of certain general rules, which may be expressed by formulae. 
In these formulae u and v will denote variable quantities, func- 
tions of x ; and c and n, constant quantities. 

It is frequently convenient to write the differential co- 
efficient of a quantity 

■ — u, instead of ■ — 
dx dx 

Thus the differential coefficient of (u -f- v) is more con- 
veniently written 

£(« + •), rather than "(" + «>. 
dx dx 

12. Formulae for Differentiation of Algebraic Functions. 



I. 


dx 


II. 


* = 0. 

dx 


III. 


d , . N du . dv 

— (u + in= 1 

dx dx dx 


IY. 


d , x du . dv 

■ — fuv) = V \-u ■ 

dx dx dx 



DIFFEEEN TIA TION. 11 



V. 


d , v du 
-_. (cu) = c— • 
dx dx 


VI. 


du dv 

V — ■ — u 

d fu\_ dx dx 

dx\v)~ v 2 


VII. 


a / n \ „ _ i au 
• — . (u ) = nu n l — • 
dx dx 



These formulae express the following general rules of dif- 
ferentiation : 

I. The differential coefficient of a variable ivitli respect to itself 
is unity. 

II. The differential coefficient of a constant is zero. 

III. The differential coefficient of the sum of two variables is 
the sum of their differential coefficients. 

IV. The differential coefficient of the product of two variables 
is the sum of the products of each variable by the differential 
coefficient of the other. 

V. The differential coefficient of the product of a constant and 
a variable is the product of the constant and the differential co- 
efficient of the variable. 

VI. The differential coefficient of a fraction is the differential 
coefficient of the numerator multiplied by the denominator minus 
the differential coefficient of the denominator multiplied by the 
numerator, this difference being divided by the square of the 
denominator. 

VII. The differential coefficient of any power of a variable is 
the product of the exponent, the power with exponent diminished 
by 1, and the differential coefficient of the variable. 

13. Derivation of Formula}. 

Proof of I. This follows immediately from the definition of 

a differential coefficient. For since — = 1, its limit — = 1. 

Ax dx 

Proof of II. A constant is a quantity whose value does 
not vary. Hence 



12 DIFFERENTIAL CALCULUS. 

Ac = and — = ; 
Ax ' 

therefore its limit — = 0. 

dx 

Proof of III. Let y = u + v, and suppose that when x is 
changed into x + h, y, u, and v become y\ u' } and v'; then 

therefore y f —y = u f —u-\-v f —v, 

that is, Ay = Au + Aw. 

Divide by Ax-, then 

Ay _ Au Av 

Ax Ax Ax 

Now suppose Ax to diminish and approach zero, and we 
have, for the limits of these fractions, 

dy __du dv 
dx dx dx 

If in this we substitute for y, u ■+■ v, we have 

d , , x du . dv 
dx dx dx 

It is evident that the same proof would apply to any 
number of variables connected by plus or minus signs. We 
should then have 

d , . , , N du .dv . duo , 
dx dx dx dx 

Proof of IY. Let y = uv; then 
y'=zu'v', 

and y'—y = u'v'— uv = (V — u)v'+u(v'—v) ; 

that is, Ay — v f Au -f- uAv. 



DIFFERENTIATION. 13 

Divide by Ax; then 

Aw ,Au . A-y 
—?-== v' \-u — • 

Ax Ax • Ax 

Now suppose Ax to approach zero, and, noticing that the 
limit of v' is v, we have 

dy du . dv 
dx dx dx 

that is, — (uv) = v \-u — 

dx dx dx 

This formula may be extended to the product of three or 
more variables. Thus we have 

— (uvw) = — (uv • w) = w — (uv) -\-uv 

dx K } dx K J dx y } dx 



( 

= wlv 



du . dv\ . dw 

— - -|- u — -\-uv 

dx dx) dx 



du , dv . dw 

= vw h uw h uv — • 

dx dx dx 

Similarly, for the product of four functions, we have 

d / x du , dv , dw , dz 

— (uvwz) = vwz 1- wzu h zuv h uvw — 

dx dx dx dx dx 

A similar relation holds for the product of any number of 
variables. 

Proof of V. This is a special case of IV., — being zero. 
But we may derive it independently thus : 

y =cu, 

y'=zcu', 

y'-y = c(u'-u), 

Ay = cAu, 

Ay _ Au 

Ax Ax' 



14 DIFFERENTIAL CALCULUS. 



dy du d , s. du 

-*- = c — , or — (cu) = c — 
dx dx dx dx 

u 
Proof of VI. Let y = -, 

then y'=-', 

v' 

,, p . w' w u'v — uv' (u'—u^v — uW—v) 

therefore y'—y = — = = ^ - — - i i; 

V V V V vv 

that is, Ay = , 



Au Av 

V u 

Ay Ax Ax 

Ax v'v 

Now suppose Ax to diminish towards zero, and, noticing 
that the limit of v' is v, we have 









du 

v 


dv 

u — 






dy 


dx 


dx 






dx 


V 2 




Or we 


may 


derive VI. 


from IV. 


thus : 


Since 
therefore 




y = 

yvz 


u 

V 

= u. 




By IV, 




dy . dv 
v— + y — = 


du 



dy du u dv 

v— — 



therefore 



dx 


dx v dx 


dy 


die dv 

V u 

dx dx 


dx 


V 2 



DIFFERENTIATION. 15 

Proof of VII. First, suppose n to be a positive integer. 



Let 


y=u n , 


id, 


y'=u'% 




y'— y = u' n — u 1 



that is, Ay = Au(u' n - 1 + u' n - 2 u + u' n - 3 u 2 htf 1-1 ), 

*y = (ti'— x + w'*- 2 ^ + % ,n - 3 w 2 ... + u^ 1 ) — . 

Az V 7 Aic 

Now let Ace diminish ; then, u being the limit of u', each 
of the n terms within the parenthesis becomes u n ~ l ; therefore 

dy n-id u 
— 2. = nit" x — 

d» dec 

Second, suppose w to be a positive fraction, — 

p 
Let 2/ = uq j 

then y*=u p -, 

therefore — («*) = — (tf ) . 

dx V ^ ; dx V y 

But we have already shown VII. to be true when the ex* 
ponent is a positive integer ; hence we may apply it to each 
member of this equation. This gives 

a -i dy __, du 

qy q 1 -f=pu p 1 — ; 
ax dx 

therefore dy = P^du_ 

dx q y q ~ x dx 
p 
Substituting for y, w, gives 

dy pu p ~ 1 du p p --\du 
dx~ q P - p -dx ~~ q dec' 

which shows VII. to be true in this case also. Hence that 
formula applies to any positive value of n, whether integral 
or fractional. 



16 DIFFERENTIAL CALCULUS. 



Third, suppose n to be negative and equal to — m. 
Let y = ir m = — ; 

(u m ) —mu m ■ — ■ 

. XTX dy dx dx ,du 

by VI., -f- = s- = 5- = — mr^V* 

J ' dx u 2m u 2m dx 

Hence VII. is universally true. 



EXAMPLES. 

Differentiate the following functions : 

1. y = x\ 

If two quantities are equal, their differential coefficients 
must be equal. Hence 

dx dx 
If we apply VII., substituting u = x and n = 4, we have 

— (x 4 ) = 4x 3 — = 4^, by I. 
dx v 7 dx ' J 

... ^/ =4**. 

2. ?/ = 3x 4 -f-4x 3 . 

1 = 1(3^+4^=1(3^ + 1(4^ 

by III., making u = 3 x 4 and v = 4x 3 . 

A(3x 4 ) = 3^-(x 4 ), by V., 
dx dx 

= 3- 4x3= 12 x 3 . 

Similarly, — (4X 3 ) = 4— (x 3 ) = 4 • 3x*= 12 x 2 . 
ax (xx 

.-. ^ = 12x 3 +12x 2 =12(x 3 +x 2 ). 
dx 



3. y = x 2 + 2. 



DIFFERENTIATION. 17 



dx dx dx 

!(*t ) = p, by yn. 

|(2) = 0, by II. 

do; 2 



4. 2/ = 3v^ — |r + - 3 + a. 



dx dx dx dx dx 

= 3 x ~i -2(-l\x-% -3x-*+0 



3 13 

2x* x* x 



k ^4-3 

5. v = — — 

y x 2 +3 

dx dx\x 2 -\-3j 
Applying VI., making 

u = x H- 3 and v = x 2 -\- 3, we have 

„,, +8N ^+8)A (a , + 8)- (a , + 8)^ +8 ) 



dxV« 2 +3y (ar+3) 

= a ?+3 -Q + 3)2^ = 3 -6x -x 2 
2 +3) 2 (^+3) 2 ' 

dy _ 3 — 6x — x 2 
dx~ (ar>+3) 2 



18 DIFFERENTIAL CALCULUS. 

6. y = (x*+2)l 



cly = d L 
dx dx 



— (a?+2) T . 



If we apply VII., making 



o 
u= x 2 + 2 and n = -, we have 

o 

|(^ + 2f=|(^ + 2p|(^ +2 ) 

2,.., , ox-lo.. ±X 



= ^(aj a +2) 3 2a = 



3 3(^ + 2)* 

(fa/ _ 4 a? 
d * 3(a; 2 +2)3 



7. y = (x 2 +l)Vx 3 -x. 

If we apply IV., making 

u = x 2 +l and v = (a; 3 — a;) 5 ', we have 

= («*•+ l)-f (a 8 - »)*+ (*>- x$±{&+ 1). 
eta! eta/ 

— (a? - a)*= - (a 8 - a?)~* — (aj 8 — ») = - (ar 5 - a)~ *(3a; 2 - 1) . 

-f (a^ + l)=2a;. 
dx 

%t r= 1 (3*+ 1) (3a; 2 - 1) (a; 3 - »)"*+ (a? 3 - »)*2» 
eta; ^ 

_ (s»+ 1) (3a; 2 - 1) + 4a;(ar 3 - x) = 7a: 4 -2ar*- 1 
2(a^-a;)^ 2(3 3 -<e)* 



DIFFERENTIATION. 19 

8. y = (x+l)\2x-l)\ j y = (16x + l)(x + l)\2x-l) 2 . 

(XX 

9 a -\- bx + ex 2 dy a 

x dx x 2 

x % dx 3 3 

-- _ x%+x — x^+a dy _ 2 x 1 — x -\- 2 x^ — 3 a 

12. Given 

(a 4- x) 5 = a 5 +5a 4 x + lOaV-f lOaV-f- 5ax 4 -\- x 5 ; 
derive by differentiation the expansion of (a -f- x)\ 

13. Given 1 + s + a? 8 ... +»"= - r^; 

£C — 1 

derive the sum of the series 1 + 2x + 3x 2 h waf 1-1 . 

rtaf+ 1 -(n + 1)^+1 
(*-l) 2 



\l-a; da; (l-aj)VI^? 

1K x n dy ?iaf _1 

15. •?/ = • -£■= • 

(l+a;) n dx (l + ») n+1 

16. y=(l-2aj + 3aj 2 -4aj 8 )(l + a0*. ^ = ~20x 3 (l + x), 

aa; 

17. 2/= (1-3^+ 6a; 4 ) (1 + a; 2 ) 3 . ^ = 60 a? 5 (1 -far ) 2 . 



20 DIFFEBENTIAL CALCULUS. 



18. y = x* (a + 3 x) 8 (a -2x)«. 



ty- = 5x\a + SxY(a-2x)(a s + 2ax-12x t ) 

CLX 



19. y = x Vi {a-ZxY{a-\-bx)K 



%L = Ux'^a-Sxy (a + 5a) 2 (a?+2ax-2Sx 2 ), 
dx 



20. y = (a + x) n (b + x)' 



ty- = [m(6+x) + n(a+x)] (a+x^'^b+x)*- 1 
dx 



21. y = 



(a + x) m (b + x)* 

dy__ m (5 + x) + n (a + x) 

dx~ (a + aj) m+1 (H«) B+1 ' 

x dy __ 1 



VT^x 1 dx (l-x 2 )* 

1 — x dy __ 1 +x 

Vl+1? <te~ (1+x 2 )*' 



24 v= 2 v* tf y = gg-^) 

* 3 + x 2 dx (3+a^)V» 



25. y = 



dy __ x_ 



-t-Vl + x 2 daj VT+x 2 " 

26. y = ?L_. 

3+Vl^ 



27. „, _ V a + a -f Va — a? 

Vo^Fa ~~ Va — x 

oq .._ 3x* + 2 



da; 


2x(l-x 2 )+Vl- 


■X 2 


dy 


a 2 -}-a Va 2 — x 2 




dx 

d y - 


x 2 Va 2 — x 2 
2 





a(a? + l)f dx x 2 (x 3 +l)f 

l. 2/ = 3(x* + l)£(4x*-3). ^=56x s (x 2 -f l)i 

dx 



DIFFERENTIATION. 21 



30. V QK + a ?. dy_ (g?-2o)Va? + tt 

V# — a ^ (a; — a) * 



32. 2/ = 



Vl + z 2 - Vl - ^ 

dy ny 



1 + Vl-«V ^ ccVl-ar 2 



14. Formulas for Differentiation of Logarithmic and Expo- 
nential Functions. 

du 

VIII. -— log a u = log a e — 
dx u 

du i 

TV d , cte ^y f 
IX. — log e w = x<< 

dx dx \ 

XL Ae--^. I 

. Xll. ■ U v — VU vl \-iOg e U'U v 

dx dx dx 

15. Before deriving these formulae it is necessary to find 
the limit of the expression 

1 -f- - ) , as z approaches infinity. 
By the Binomial Theorem 

■ ,(,-iH,-2) m V 



h may be written 



22 DIFFEBENTIAL CALCULUS. 



Now when z increases indefinitely, we have 

i+i 

iri» 

This quantity is usually denoted by e, so that 



limit of /l + iY=l + l 



e=i+ i + H + "" 

The value of e can be easily calculated to any desired 
number of decimals by computing the values of the successive 
terms of this series. For seven decimal places the calculatioD 

is as follows, — 

1. 
1. 
.5 

.166666667 
.041666667 
.008333333 
.001388889 
.000198413 
.000024802 
.000002756 
.000000276 
.000000025 
.000000002 



e = 2.7182818... 

By calculating the value of f 1 + -J for different values of 2, 
we may verify its limit. Thus 

(1+ iY =2.25 

(l + iy=2.48832 

(1+^)10=2.59374 

(1.01) 100 = 2.70481 

(1.001) 1000 = 2.71692 

(1.0001) loaeo = 2.71815 

(1.00001) 100000 = 2.71827 

(1.000001) 1000000 = 2.71828 



DIFFEBEN TIA TION. 23 



16. Derivation of Formulas. 
Proof of VIII. Let y = log a u, 
then y'= log a (u + Au), 



Ay = log a (u + Au) — log a u = log a - - 



Su 



Dividing by Ax, u Au 

Ay . / AuY" Ax 

Now if Ax approach zero, Au at the same time approaches 

u 

zero ; then the limit of ( 1 -\ J u is the same as the limit 

/ IV • 

of ( 1 -f- - ) as z increases indefinitely. Bnt in Art. 15 we 

have already found the latter limit to be e. Hence we have 

du 
dy , dx 

Proof of IX. This is a special case of VIII., when a = e. 

In this case , . , „ -. 

log a e = log e e = 1. 

Note. — Logarithms to base e are called Napierian loga- 
rithms. Hereafter, when no base is specified, Napierian 
logarithms are to be understood. 

That is log u = log e u. 

Proof of X. 

Let y = a u . , 

Taking the logarithm of each member, we have 

logy =u log a; 
dy 

therefore by IX., ^.= Wa- . 
y dx 



24 DIFFERENTIAL CALCULUS. 

Multiplying by y = a u , we have 

dy -, u du 

-*- = \oga>a u — 
dx dx 

Proof of XL This is a special case of X., where a = e. 

Proof of XII. Let y = u v . 

Taking the logarithm of each member, we have 

logy = vlogu ; 

_£ v — 
therefore by IX., dx _ da? , i ~^. 
y u * dx 

Multiplying by 2/ = ^ v > we have 

dy v ,du . -, „cZy 

dx dx dx 



EXAMPLES. 

2. y = xlogx. -|=l + loga>. 

3. 2/ = a; n logaj. -^ = a; n - 1 (l + nlog»), 

dy 



4. ^logVl-a 2 . ^ = -w- 

5. y = e*(l-x% | = ^(l-3^-^). 

6. 2/= ya; -iog(v» + l). 

7. 2/ = log (log a). 



fy _ 


1 




dx 
dy_ 


2(V* 
1 

xlogx 
e 2x -l 


+ 1) 



DIFFERENTIATION. 25 

9. 2/ = (a;-3)e 2 * + 4a;e* + x. ^■ = (2x-5)e* a +4;(x+l)e*+l. 

dx 

where M = — — = log 10 e = .434294 



11. t/ = 5 a:8+2a? . 



12. ? = *■ 



e x + e~* 





log. 10 ~*»~ 




da; - 


= 2(aj + l)5* s + 2 * 


log5, 




log 5 = 


; 1.609440 


da; 


4 


, 



What is the result of differentiating both members of each 
of the three following equations? 

13. log(l+ a! ) = a! -| + |-| 4 +- 

Ans. = 1 — aj-h^c 2 — »*+ ••• 

1 -f-a; 



5 l-a; V35 7 / 



.4ra«. — i— = l+ar ! +a; 4 + a^ + 
1 — ar 

15 . ,. 1+ . + | + | + g + ... 

Ans. e-=l + , + | + | + ... 

16. y = x n a a . ( ^ = x n - l a x (n + x\oga). 

dx 

* BV ; (a;-2) 2 da; (a;-2) 8 

18. y-l g V CT + V g ^ = _v^L_. 

■Y/a — v'x da; (a — a;) y/x 



= e 






26 

19. y 

20. y 

21. y 

22. y 

23. y 

24. y 

25. i/ 

26. y 

27. y 

28. 2/ 

29. 2/ 

30. y 

31. 2/ 

32. y = 



DIFFERENTIAL CALCULUS. 



fcloga; 



log(l-z). 



1-x 



dy_ logs 

d»"" (i — a?) 2 .* 

da; 2 



|[(log^) 2 -logV^ + 4]. ^ = ^(lo ga; ) ! 

\ a a 2 ay 

log a; • log (log a;) — logo;. 



dx 
dy 


= aa^e *. 
log (logs?) 


dx 


X 



log(a>-8 + Va; 2 -6a;+13). ^= * 

dx Var*-6a; + 13 



= mlog ( ^/x + VaT-j-m) + Vmx + x*. 

dy 
dx = yj 



m + x 



= log 



= log 



X 



a?V2 + Vl+ar 8 dy y2 



Vf=a? 



da; (i-{r*)VlT? 



_ , V a? + a 2 + Var* + & 2 dy _ 



2a; 



Va^ + a 2 - VaF+tf dx Vx T Ta 1 Vx Y +b 2 



, b-l , . la 3 -fl 
= l0g >|^Pl +l0g \^l 



dy_ a^ — 1 
da; aj 4 -j-ar*+l 



( e ._ e -*)2 (e 2 ^ 2e 4 *+ 3e 6 *) . ^ = 24e 6 * (e 2 * - 1) . 

dx 



a^. 



#*« l-2x 

— = aj - (1-loga;). 

2-«©"( ,+ "<} 







DIFFERENTIATION. 27 

33. y = (ex)*. g = (eaj)-'(2+log«). 

35. v = a log *. ^=log^.a; losa: - 1 . 

die 

36. y = a^ ^ = 0. 
9 dx 

37. v=e eX . \ - ^ = e e V. 

38. v = e^. ^ = c^af(l+ logos). 

39. 2/ = ^. ^=yx x \- + l °g x +( l °g x ) 2 \ 



17. Formula? for Differentiation of Trigonometric Functions. 
In the following formulae the angle u is supposed to be ex- 
pressed in circular measure. 

XIII. — sin^=cosw— • 
dx dx 

XIY. — cosw = — sin it— • 
dx dx 

\TTT VV i O iljW 

XV. — tanw=sec J w — 
dx dx 

XVI. — cot u = — cosec 2 % — 
dx dx 

XVII. ■ — sectt = secit tanw — 
dx dx 

XVIII. — cosec w = — cosec u cot u — • 
dx • dx 

XIX. — vers u = sin u — 
dx dx 



28 DIFFERENTIAL CALCULUS. 

18. Derivation of Formulae,. 

Proof of XIII. Let y = sinw, 
then y' = sin ( u 4- Aw) ; 

therefore Ay == sin (w + Ait) — sin u. 

But from Trigonometry, 

sin .4 - sin£ = 2sin-(J. -B) cosi(J. + £). 

.4 w 



If we substitute 


A = u-\- Au and B = u, 


we have 


Ay = 2 0OS (u + fyi a f. 


Hence 


. Aw 
sin — 
Aw / . Aw\ 2 Ait 

— s. = cos [ w H ] 

Aa; \ 2 J Au Ax 




2 



Now when Ax approaches zero, Aw likewise approaches 
zero, and as Aw is in circular measure, the limit of 





. Au 
sin — 

is unity. 

Aw J 




2 


Hence 


dy du 
-£= cosw — 
dx dx 


Proof of XIV. 


This may be derive 


XIII. for w, - - w. 

' 2 





Then A sin ^ - u) = cos (^ - u) - (?- - u\ 
dx \2 J \2 Jdx\2 / 



DIFFEBENTIATION. 29 

/ du\ ■ < 

f =-smw 

\ dx) i 



d f du\ • du 

or — cosw = smw =-sm« — 

dx \ dx) dx 



n „ . „„ _. . sin w 

Proof of XV. Since tan w = , 

^ ^ cos u 



d . d 

cos ?* — sin w — sin w — cos w 
d , da; da; 

by VI., — tan u = 

dx cos J w 



2 du , . 2 du du 
dx dx dx 



cos'u cos'u 

o du 

= sec J % — 

dx 

Proof of XVI. This may be derived from XV. by substitut- 
ing - — u for u. 

Proof of XVII. Since sec u = 



by VI., 









cosw 






d 


secu 


= 


d 

— - — cos^ 
dx 


i 


du 

unit — 

da; 


dx 


GOS 2 U 


COS 2 W 






= 


sec u tan u 


du 
dx 





Proof of XVIII. This may be derived from XVII. by sub- 
stituting - — u for u. 



9 



Proof of XIX. This is readily obtained from XIV. by the 
relation 

vers it = 1 — cosw. 

. 



30 DIFFERENTIAL CALCULUS. 



EXAMPLES. 

1. y = sin 2 x cos x. -^ = 2 cos 2 x cos a; — sin 2 x sin a> 

da; 

2. y — tan 2 5a;. -^= 10tan5a;sec 2 5a;. 

da; 

3. y=: tana; — x. -^ = tan 2 a;. 

da; 

4. y = sm(nx + m) -^ = n cos (nx + m). 

dx 

K tana; — 1 dy 

5. y = • — = sina;-f-cosa;. 

sec a; dx 

6. y = sin 3 a; cos a;, -f- = sin* x (3 cos 8 x — sin 2 a;). 

da; 

7. y = sin(a;-}-a)cos(a; — a). -^ = cos 2a;. 

da; 

sin (a — a;) dv . ., N 

8. y = ^ '-' _£ = — sin2acosecr(a + a;), 

sin (a + x) dx 

9. y = tan 2 a;-log(sec 2 a;). ^=2tan 8 a;. 

(XX 

10. 3/ = tan 4 x— 2 tan 2 a; + log (sec 4 x). 

^ = 4tan 5 a;. 
da; 

11. y = (asin 2 a; + &cos 2 a;) n . 

-^ = n (a — 6) sin 2 x(a sin 2 a; •+- 6 cos'a;)"" l , 
dx 

12. ?/ = log sin x. -^=cota;. 

eta; 

13. y — log tana;. 



da; sin 2 a; 



14. y = log sec a;. -^ = tana;. 

da; 



DIFFERENTIATION. 31 



15. y = vers(- + a; Jversj ^ — x\ 



-^ = — sin 2 x, 
dx 

16 e-fosins-cosaQ. ^ = e «. sina . # 

* a 2 4- 1 da; 

-- .;._ dv /sina; . \ 

17. y = x mm . 3^ = 2/ h cos a; log a; • 

da; V * * y 

18. y = sinna;sin n a;. -^ = nsin w ~ 1 a;sin(n-|-l)a;. 

da; 

,q __ sin m yia; dy _ mn sin m " 1 yia; cos(m — n) x 

cos n mx dx cos n+1 mx 



20. y = x -flog cos [a; — -]. 

21. jr = logtang + l\ 

22. ,-fcgJ}=22S?. & 

\ 1 + cos x dx 



dx 1-f-tana? 

-^ = seca;. 
dx 



-^ = coseca;. 
-hcosa; 

23 « = Iop- | ac osa; — 6 sin a; dy — ab 

\ a cos x -\-b sin x dx a 2 cos 2 a; — b 2 sin 2 a; 

o* „, tana; — tan 3 a; dv . 

24. i/ = -. -^ = cos4a;. 

sec 4 a; da; 

In each of the following pairs of equations derive by differen* 
tiation each of the two equations from the other : 

25. sin 2x = 2 sin x cos x, 
cos 2 x = cos 2 a; — sin 2 a;. 
2 tan a; 



26. sin 2 a; = 
cos 2 a; 



1 -f tan 2 a; 
1 — tan 2 a; 



1 4- tan 2 a; 



32 DIFFERENTIAL CALCULUS. 

27. sin 3 x = 3 sin x — 4 sin 3 x, 
cos3# = 4cos 3 # — 3cos#. 

28. sin 4:X = 4 sin x cos 3 x — 4cos#sin 3 # ? 
cos 4 x = 1 — 8 sin 2 a? cos 2 x. 

29. sin(m -fn)# = sin race cos ?i£ + cos m# sin ncc, 
cos (m + n) & = cos mx cos n# — sin mx sin n#. 

/y»3 />■»*> /W 

30. sina; = a-- + — - — H 

[3 [6 [7 

cos , = 1 __ + ___ + .... 

Q1 • e^- 1 - e— /=i 

31. sinas = 



cos# = 



2V^1 ' 



c *^=i + e-**^ 



19. Formulae for Differentiation of Inverse Trigonometric 
Functions. 

du 



XX. 



XXI. 



XXII. 



XXIII. 



XXIV 



d . _j dx 



dx 




VI - u 2 


d 

COS" 


*u = 


du 
dx 


dx 


VI -u 2 


-—tan" 
dx 


~ l u--^ 


du 

dx 
1 + u 2 ' 


d , 
--cot 
dx 


i u — 


du 

dx 




1 + u 2 


d 

-—see" 


1 u = 


du 

dx 



dx U -y/tf 



DIFFERENTIATION. 33 



du 



"VVT7 ^* 1 ^**^ 



dx u-Vu 2 — l 

du 

XXYI. A vftrs -i M= dx . 



20. Derivation of Formulae. 
Proof of XX. Let y = sin _1 w ; 
therefore smy = u. 

By XIII, cos2/^ = ^; 

dx dx 



therefore 



dw 
c?2/ dx 



dx cos ?/ 



But cos y = Vl — sii?y = Vl — w 2 

du 
d?/ do? 



therefore 



dx Vl 



Proof of XXI. This may be derived like XX. ? or from 

the relation 

cos -1 w == sin _1 w ; 

2 du 

, d _i d . , dx 

whence —cos l u = — -sm -I w = 



dx dx Vl — 

Proof of XXII. Let ?/ = tan- 1 w; 

therefore tan?/ = u. 

t^ VTT o dv dw 

By XV., ^yf x = a X ' 

du 

., « dv dx 

therefore -^ = — 5-. 

dx sec 2 y 



34 DIFFERENTIAL CALCULUS. 

But sec 2 ?/ = l + tan 2 ?/ = l-f % 2 ; 

du 



therefore 



dy dx 



dx 1 + u 2 



Proof of XXIII. This may be derived like XXII., or from 
the relation 

cot -1 u — - — tan -1 u. 

2 

Proof of XXIV. Let y = sec" 1 ?* ; 
therefore sec y — u. 

ByXVIL, sec?/ tan 3 = ^; 

dx dx 

du 

, o dy dx 

therefore 



dx sec y tan?/ 



But secy tan?/ = sec?/ V sec 2 ?/ — 1 = u VV — 1 ; 

du 

,, ~ dy dx 
therefore -^- = . 

d% u yV — 1 
Proof of XXY. This may be derived like XXIV., or from 
the relation 

cosec" 1 u = - — sec -1 u. 

Proof of XXVI. Let ?/ = vers- 1 u; 
therefore w = vers y = 1 — cos ?/. 

, , p d?/ da; 

therefore ~^- = — 

dec sin?/ 



But sin?/ = Vl — cos 2 ?/ = Vl — (1 — u) 2 = V 2 % — w 2 ; 

therefore ^ = _^2 

<** V2tt-u a 



DIFFERENTIATION. 35 



EXAMPLES. 



i 4. -l dy m 

da; l + m 2 ar 



2. 2/ = sin' 1 (3a; — 1). 

3. y = vers 



-i i^. ^ — 2 

9 dx V9a?-4ic 2 

4. y = sin- 1 (3a;-4ic 3 ). ^= 8 - 

da> VT^ 

5. 2/==t an- 1 -^-. ^ = _1_. 
* 1-ar* d» l + a? 

6. 2/ = tan-V. ^ = _J_. 

da; e"+ e— 



7. y = tan" 1 (n tan a;) 



dy n 



dx cos 2 x-^n 2 sin a x 



8. y ^cosec" 1 — - %L 



2 



2x dx V9 — 4^ 

9. y= versus 1 . ^= 2 ■ 

1A _i 2CC 2 dw 2 

10. v = vers * -• -£ = -. 

9 1 + x 2 dx l+x 2 

11. y.^tC. % = _!_. 

2 da; e*+e-* 

12. 2 , = cosec- 1 -^— • JJU 2 * 

2a a -l da; VI^^ 

Itj. v = sec 1 -r- ! — • -£ = — - . 

9 x*-l dx ar*+l 



36 

> 14. y 

15. y 

16. y 
7 17. y 

18. y 

19. 2/ 

20. y 

21. 2/ 

22. y 



DIFFERENTIAL CALCULUS, 



= sin" 1 



-ifc + 1 



V2 



= tan 



= cos 



_! 4 sin a; 
3+5 cos a; 

.j 3 + 5 cos a; 
5 + 3 cos a; 



sin x -• 

1+ar 2 



= cosec 



tan 



-il+s 2 
2a; 

_! a; + a 
1 — oa; 



sin" 1 Vsina; 
tan 



-tan-J* 



cos a; 



+ cosa; 



tan 



_! Va; + Va 



1 — -Vax 



cot-" + log J^. 
a; \a; + a 



24. y 

25. y 

26. y 

27. y 



tan-Xaj+Vl-a^). 
!€* — e— 



= cos 



^■^ con~^ 



e* + e" x 
2" 



^_ 


1 


c?a; 


Vl-2a;-a? 


<fy_ 


4 


da; 


5 + 3 cos x 


%_ 


4 


dx 


5 + 3cosx 


da; 


-2 

1+ar 2 


<ty_ 


2 


da; 


1+a; 2 


%_ 


1 


da; 


1+ar 8 

i .... 



-^ = - V 1 + cosec x, 

dx 2 

da; 2 

dy_ 1 

da; 2V^(l + ») 

dy _ 2 ax 2 
dx a; 4 — a 4 



Vl-ar>- 



d?/_ 

<^ ~ 2 VT^sFCl + a; Vl^aJ 8 ) 

dy = - 2 

da; e^e"* 

d2/___-l 



1+a; 



da; 2Vl-a? 

(aj+^tan" 1 i-VS. ^ = tan" 1 §. 
\a da; \a 



DIFFEREN TIA TION. 37 



28 . y = cot -i l + Vl + * . Jr 1 

a; da? 2(1 + or) 

oft . , xtano. d?/ a 2 tana 

29. y=sin- 1 — :£=^ — la- 

va 2 — ce 2 



30. y = cot- 1 Ji^^. 

* \2 + a> 

31. y = tan- l 3a? "" g; > - 
5 1-3^ 

32. y^tan- 1 2x , + cot- 1 

* 1 + 3* 2 

33. y = t3LK- l 2x ~_ b + tan- 1 - 

6V3 a?V3 ^ 



dec 


a 2 - 


«" vV - 


a; 2 sec 2 a 


%_ 




1 




dx 


2Vz 


5 — x — x 2 




dy _ 


3 






dx 


1 + - 


c 2 




i X 
l + 2x 2 


-f tan - 


>2o, ^- 
dx 


3 

l + 9a 2 


2b-x 




dy = 


:0. 



34. 



y = losr \ - {- tan -1 

3 *V2x 2 + 2x+-l \-2x 2 



dy Sx 2 



dx 4.x* + 1 



21. To express JL in terms of — If y is a function of x, 
dx dy 

then (Art. 2) x may be regarded as a function of y. From 

the former relation we have JL, and from the latter, — 

dx dy 

These differential coefficients are connected by a simple relation. 

It is evident that _ ^ = — , 

Ax Ax 

Ay 

however small the values of Ax and Ay. As these quantities 

approach zero, we have ? for the limits of the members of this^ 

equation, 

^=1 a) 

dx dx 
dy 

That is, the relation between _£ and — is the same as if they 
were ordinary fractions. ^ 



38 DIFFEBENTIAL CALCULUS. 

For example, suppose 
a 



y + l 

Differentiating with, respect to y, we have 
dx a 



(2) 



<fy (2/ + 1) 



This is the same result that we get by solving (2) with 
reference to y, giving 

a . 

and differentiating this with reference to x. 

22. To express -JL in terms of -z. and — If y is a given 
dx dz dx 

function of z, and z a given function of x, it follows that 

y is a function of x. This relation may be obtained by 

eliminating z between the two given equations, but ~M. can be 

dx 

found without such elimination. 

By differentiating the two given equations, we find -Jl and 

_, and from these differential coefficients, -^ may be obtained 
dx dx 

by a relation which may be derived as follows : 

It is evident that — = — — , 

Ace Az Ax 

however small Ax, Ay, and Az. As these quantities approach 
zero, we have for the limits of the members of this equation, 

^/ = <fy^. /-|N 

dx dz dx' 

That is, the relation is the same as if the differential coeffi- 
cients were ordinary fractions. 



DIFFEBEN TIA TION. 39 

For example, suppose 

* = *l A (2) 

z = a 2 — or. ) 

Differentiating these equations, the first with reference to z, 
and the second with reference to x, we have 

dz dx 

By (1), ^- = 5z\-2x) = -10x(a 2 -x 2 Y, by (2). 

CLX 

The same result might have been obtained by eliminating z 
between (2), giving 

y = (a 2 - x 2 ) 5 , 

and differentiating this with reference to x. 



EXAMPLES. 

ffqc 

In the following seven examples find by differentiation — , 
7 ty 

and then ?0L by (1) Art. 21. 
dx 



da? 2 (a? - 2) 2 



2. o;=V2/ 2 4-l- 



<^y _ vV + 1 _ _ ^ 2 + 1 
dx y-s/7 + 1 2 ^ ' 



o /T~n — : ^V ^ V 1 -h sin V 2 

3. »= Vl + siny. -^ = — - — ' ^= — ■ 

dx cos y ^/2 — x 2 

4. x = tarr 1 (y + -Vy 2 -l). ^ = 22/V^ = l=l(tan 2 ic-cot 2 a;). 

5. ^_ 2/ */_(l+log?/) 2 _ 2/ 2 



1+logy dx log y xy — x 1 



„ , e y + V e % - 4 cty V e 2 * — 4 e 2 * - 1 

2 dx e y e 2x + l 



40 DIFFERENTIAL CALCULUS. 

vV + 2 + vV — 2 dy Ve 2 * - 4 e 2 * — 1 



7. x = 2log- 



dx e y e 2x + 1" 



In the following examples find by differentiation _? and — , 
^ cfc dec 

and then ^ by (1) Art. 22. 

2z _ x dy 4 

9. ?/ = e*-f-e 2z ; 2 = log (x — x 2 ) . 

10. y — log (z* — z), z = e 3 *. 

11 i I + 2 2 

11 . ?/ = log — — , 2! = e x . 

z 

12. ?/ = tan2z, z = tan" 1 (2x - 1). 



13. y-llog (^^-Xta^^ZLl, ,^1+32 + 3* 
* 6 V-* + l V3 V3 ' » 



dx 


(X- 


■2) 2 " 


dy_ 
dx 


= 4^- 


-6a* + l. 


dy 


5e 2x 


-3 


dx 


e 2x _ 


-1 


dy 


e x — 


e~ x 


dx 


e x + 


e~ x 


dy 


2x 2 - 


-2* + l 


dx 


2( 


x-x 2 ) 2 



dx xz (1 + a;) 



CHAPTER IV. 

SUCCESSIVE DIFFERENTIATION. 

23. Definition. A single differentiation performed on 

y=f(x) gives the differential coefficient, -^« This result 

dx 

being generally also a function of x, may be again differen- 
tiated, and we thus obtain what is called the second differential 
coefficient; the result of three successive differentiations is 
the third differential coefficient; and so on. 

For example, if y = a? 4 , 

dx 
dxdx 

dxdxdx 

24. Notation. The second differential coefficient of y with 

d 2 v 

respect to x, is denoted by — ^- 

That is, 
Similarly, 



cVy = 
dx 2 


d dy 
dxdx 




d 3 y _ 
da7 3 ~" 


d d dy 
dxdxdx 


_d_dty t 

dx dx 2 


die 4 


d d d dy _ 
dxdx dxdx 


d d B y 
dxdx s 


... 


... 


... 


d^_ 


_ d d n ~ l y 




dx n ~ 


dxdx"- 1 





42 DIFFERENTIAL CALCULUS. 

Thus, if y = x 4 , 

dsc 

dor 

^ = 24*. 
da; 3 

The successive differential coefficients are sometimes called 
ihejlrst, second, third, • •• derivatives. 

If the original function of x is denoted by f(x), its suc- 
cessive differential coefficients are often denoted by 

/'(»), /"(*), /"'(*), - /"(*); 

25. 27ie w^ Differential Coefficient. It is possible to express 
the nth. differential coefficient of some functions. 
Tor example, 

(a). From y =e x , we have 



dy = e* ^-=e* ... ^ = e* 
da ' dx 2 ' dx 11 



(b). From y = e ax , we have 

da; dor da n 

(c) . From 2/ = log sb, we have 

dX 4 V ' L_ > dx n x n 



SUCCESSIVE DIFFERENTIATION. 43 

(d) . From y = sin ax, we have 

-¥■ = a cos ax — a sin [ ax 4- - ], 
dx \ 2/ 

J = a 2 cos( ax + - ]= a 2 sin[ ax + — }, 
dx 2 \ 2J \ 2 J 



d s y 3 / , 2tt 

— £ = a 3 cos [ ax H 

dx 3 V 2 



— £ = a n smf ax-\ ]• 

dx n \2J 



EXAMPLES. 



1. 2/ = x 4 -4x 3 +6x 2 -4x + l. ^ = 12(x 2 - 2x + l). 

c?x 2 

2. y = rf. § = £• 



3. 2/ = (x-3)e 2x +4xe w +x. ^= 4e*[(x- 2)e x +x-f 2]. 

dx 2 

» _ a d 2 y _ m(m + l)a 

V ~x™' 



5. 2/ = xlogx. 

6. t/ = x 3 logx. 

7. 2/ = log(e a; +e- a: ). 



dx 2 




x m+2 




d?y_ 
dx 2 


_1 

X 






d 4 y_ 
dx 4 


_6 # 

X 






d 3 y _ 


= — 


8(e*-e 


-*). 


dx 3 " 


(e^+e- 


o 3 



44 DIFFERENTIAL CALCULUS. 



8. y = (x*-6x+12)e x . ^| = rcV. 

(iXt 

9. y — — [logo; )• — \ — x log x, 

9 6\ 8 6J dx 2 s 

, n t . d 3 y 2cosa; 

10. 2/ = logsina\ — ,= 

dor sin 3 a; 



11. ^ = (ic 2 +a 2 )tan- 1 



_! # d 3 ?/_ 4a ; 



a daj 3 (a 2 -+- x 2 ) 

d*y 



12. 2/ = e"*cos#. — ^ = — 4e'*cosa;. 

do; 4 

13. y = tana;. — ^ = 6 sec 4 a; — 4 sec 2 a;. 

dx 3 

i4. y =^±i. ^ =L6 r^_ + 2 -] 

* ar*-l dx 6 L -[_( iC -l) 7 (a;+l) 7 J 

Decompose the fraction before differentiating. 

15. 2/ = ysec2a;. |^=3^-?/. 

16. y = {e+e- x ) n . ^| = n 2 2/-4n(n-l)^. 

dar 

.,« 7 cos a; cos 8 a; d z y . « 

17. y = — ^- = sin 3 a;. 

9 9 27 (to 3 

18. v= tan 2 a; + 8 log cos x + 3 a; 2 . ^-| = 6tan 4 a;. 

dar 

19. y = (x i -dx + 3)e ix . ^l = 8x 2 e i *> 

OLX> 

20. 2/ = a^r3(log i r) 2 -llloga ; + ^"]. 0= 18 (log a;) 8 . 

21. v = e aa: sin6a;. -^|-2a^ + (a 2 + 6 2 )v = 0. 

dx 2 dx 

22. y = sin(msin- 1 a;). (l-a; 2 )^- ~-x^- + m 2 y = 0. 

dx 2 dx 



SUCCESSIVE DIFFERENTIATION. 45 

23. y — a cos (logo;) -\-b sin (logic). <c 2 ^| + a;-^ + ?/ = 0. 

(XX" (XX 



24. y 



1 d H y = (-1)"^ 

ic + 2* dx n (a> + 2) n+1 ' 

25 V=± _I_. d^^ (-l)"3»ln 

- 3a5 + 4 dx n (3aj + 4) n+1 



26. y=a*. ff = (log a)»a*. 

dec" 



27. v = cos aa?. 



^ = a w cos aec H ]■ 



^ 1 + sc cfo n (1-M) n+1 

2 
Keduce the fraction to a mixed quantity, — 1 H , before 

differentiating. ' 



29. y = 5|+^. *S = (-1)-| W 

J a 2 - 4 dx n v y L 



9 



(a + 2) n+1 (cc-2) w - 
«, 

/».'' r . ' . 

26. Leibnitz's Theorem. This is a formula for the wth 
differential coefficient of the product of two variables in terms 
of the successive differential coefficients of those variables. 

A special case of Leibnitz's Theorem, when n = 1, is formula 

IV, 

d , x du . dv /ix 

— (uv) = — v-\-u — (1) 

dx dx dx 

For convenience let us use the following abridged notation : 

dv d 2 v d n v 

v i = —? v 2 = — -», '" v n— 

dx dx 2 dx n 

_ du _ d 2 u _ d n u 

1 — ' ? ^2 — o? *** ^n — * 

dx dxr dx n 

Then (1) becomes 

— (uv) = u 1 v + uv l (2) 

dx 



46 DIFFERENTIAL CALCULUS. 

Differentiating (2), 

d 2 
— - 2 {uv) — u 2 v + i^i+ ^1^+ uv 2 = u 2 v -f- 2u 1 v 1 -\- uv 2 . 

(XX" 

d 3 

— -(uv) = u 3 v -+- u 2 v 1 +2uoV 1 -\-2u l v 2 + UjV 2 + uv 3 
dx 3 

= U 3 V -J- 3 U 2 V± + 3 U ± V 2 -\- UV 3- 

We shall find this law of the terms to apply, however fai 
we continue the differentiation, the coefficients being those of 
the Binomial Theorem. 

In general 

— (uv) = u n v + nu n _ 1 v 1 + v J u n _ 2 v 2 + ••• 

+ nu 1 v n _ l +uv n (3) 

This may be proved by induction, by showing that if true for 

d n • d n+l 

— (uv), it is also true for -Auv). This exercise is left for 

dx nK n dx n+lK J 

the student. 

In the ordinary notation (3) becomes 

d n , N d n u . d n ~ 1 udv , n(n — l)d n - 2 ud 2 v , 

. — (uv) = — v -\-n — H — ^ '- — --\ 

dx nK J dx n dx^dx [2 dx n ~ 2 dx 2 



dud n ~ 1 v , d n v 

. + u — 

dx dx 71 - 1 dx n 



+ n 7+u — (4) 

-7 i-„n— 1 -7..M 



For example, let us find by Leibnitz's Theorem — (e^x] 

cix 

Here u=e ax , u 1 =ae ax , ••• u n = a n e ax . 

v = x, v ± =l, v 2 =0, v 3 = 0, •••. 

Substituting in (3), we have 

^—(xe ax ) = a n e ax x -f- na n ~ 1 e ax = a^e^tax -f n). 
dx n 



SUCCESSIVE DIFFERENTIATION. 47 



EXAMPLES. 

Find by Leibnitz's Theorem the following differential co- 
efficients : 

1. y = x 3 t&io L x. — - = 2x? sec 2 cc(3tan 2 # + 1) + 18 x 2 sec 2 x tana? 

dx 3 

+ 18 x sec 2 x -f- 6 tan cc. 

2. 2/ = e'log*. - = e^log* + _-_ + _- _j. 

d n v 

3. y=x 2 a x . — - = a x (loga) n_2 [(a;loga + n) 2 — n]. 

3 (x + lf dx n K J L (£C + l) n+3 



CHAPTER V. 

DIFFERENTIALS. 

27. The differential coefficient J^. has been defined, not as a 

dx 

fraction having a numerator and denominator, but as a single 

symbol representing the limiting value of — , as Ax and Ay 

approach zero. But there are some advantages in regarding 
the differential coefficient as an actual fraction, dx and dy 
being infinitely small increments of x and y, and called differ- 
entials of x and y. That is, dx is an infinitely small Ax, and 
dy an infinitely small Ay. 

Tor instance, if we differentiate y = x 2 , we obtain 

< fy = 2x. 
dx 

Using differentials, this result might be written 
dy = 2 xdx. 

These are two forms of expressing the same relation. Ac- 
cording to the first, — 

The limit of the ratio of the increment of y to that of x, as 
these increments approach zero, is 2x. 

According to the second, — 

An infinitely small increment of y is 2 x times the corresponding 
infinitely small increment of x. 

We have the same two forms of expressing other relations 
in mathematics. 

Tor instance, we may say, — 

" The limit of the ratio, ■ , as these quantities approach 

., „ chord 

zero, is unity," 



DIFFERENTIALS. 49 

Or,- 

" An infinitely small arc is eqnal to its chord." 

The equation dy = 2xdx may thus be used as a convenient 
substitute for , 

-^- = 2x. 
ax 

We see also why -^ or 2 a? is called the differential coefficient, 
clx 

for it is the coefficient of dx in the equation dy = 2xdx. 

28. The formulae for differentiation may be expressed in 
the form of differentials by omitting the dx in each member. 
Thus, IV. becomes 

d(uv) = vdu + udv ; 



and XXII., dtan~ 1 w = - 

l + iu 



2' 



and the others may be similarly expressed. 

Differentiation by the new formulae is substantially the 
same as by the old, differing only in using the symbol d 

instead of — 
dx 

For example, take Ex. 5, p. 17. 

dv = df*+J*\ = {X " + 3)d(a + 3) -(x + 3)d(x 2 + 3) 
y \x 2 + 3j (V-f3) 2 

= (x 2 + 3) dx - (x + 3) 2xdx 
(z 2 + 3) 2 

= (x 2 +3-2x 2 -6x)dx = (3-6x-x 2 )dx 
(x 2 +3) 2 (»2+3) 2 

Dividing by dx gives 

dy _ 3 — 6 x — x 2 
dx (a 2 -f-3) 2 



50 DIFFERENTIAL CALCULUS. 

29. Successive Differentials. Successive differential coeffi- 
cients, — %. —V, • ••, which have been defined as single symbols, 
dx 2 dx 3 

may also be interpreted as fractions, the numerators, d 2 y, d 3 y, 
• ••, denoting d(dy), d[d(dy)~], • ••, and called the second, 
third, •••, differentials of y, while the denominators are (dx) 2 , 
[dx) 3 , .... 
This will be better understood from an example. 

Let y = x A , 

then dy = 4 x 3 dx. 

As Ax 3 dx is a variable, dy is a variable, and may be 
again differentiated. Now, x being the independent variable, 
its increment dx may be supposed the same infinitely small 
quantity for all values of x ; that is, we may regard dx as 
constant in the preceding equation. Thus we obtain 

d{dy) = 12x 2 dx . dx = 12a^(dx) 2 . 
Denoting d(dy) by d 2 y, 

d 2 y = 12 x?(dx) 2 . 
Differentiating again, and still regarding dx as constant, 

d(d 2 y) = 2±xdx(dx) 2 = 2±x(dx) 3 , 

or d 3 y = 24:x(dx) 3 . 

From these equations, by dividing by the power of dx in the 
second members, we find 

d 2 y 



(dx) 
d 3 y 



- = 12z 2 , 



= 24o?. 



(dx) 3 

The independent variable x, whose differential is supposed 
constant, is sometimes called the equicrescent variable. 



DIFFERENTIALS. 51 

EXAMPLES. 

Differentiate the following, using differentials in the process: 

- x 2 +2 , x 2 +2x-2, 

1. V = — 7-r- dy = - J ^ x2 dx. 



x + 1 (» + l) 



2z 



2 

1-w 



2. ?/=Va 2 +cc 2 . dy = — (a 2 -f-# 2 ) B da;. 



n 



3. y = (e x + e" x ) 2 . cfy = 2(e 2x - e- 2x )cte. 

4. y = e x log#. dy = e x ( logx + - jda?. 



K e x — e~ x , fe x —e~ x \ 2 -. 
5. y = # — • «y = [ — " x - 



e x +e~ x yf+ e ~ 

6. y== sin w #cos n #. cfa/ = sin m-1 a;cos n_1 a;(mcos 2 a! — nsin. 2 x)dx. 

7. ?/ = -tan 3 x + tana;. dy = sec 4 xdx. 

o 

8. ?/ = tan- 1 logcc. dy 



<1 + (logo;) 2 ] 



jj'/vv- \A. 



CHAPTER VI. 

IMPLICIT FUNCTIONS. (See also Art. 67.) 

30. Hitherto, in finding -JL, _^, _JL • ••, v has been an explicit 

function of oj. When the relation between y and a; is given by 
an equation containing these quantities but not solved with 
reference to y, y is said to be an implicit function of x. 

If the equation can be solved with reference to y, we 
may find its differential coefficients by the methods already 
given. But this solution is not necessary for the differentiation, 
for by the use of the formulae of differentiation we may derive 

-^, — , — si, •••, directly from the given equation. 
dx dx 2 dx? 

31. For example, suppose the relation between y and x to 
be given by the equation 

a y+ tfa?= a 2 b 2 . 
Differentiating with respect to x, 

— (a 2 y 2 +b 2 x 2 ) = 0, 
dx K * } 

' 2a 2 y^ + 2h 2 x = 0, 
dx 

dy _ _ Irx 
dx d 2 y 

Having thus obtained the first differential coefficient, we 
may, by differentiating again, derive the second differential 
coefficient. 

ahitf-tfxa 2 ^- b 2 (y-x c ^- 
d?y__db 2 x_ dx_ \ d x 

dx*~ ~ dxa 2 y~~ a A y 2 a 2 y 2 



IMPLICIT FUNCTIONS. 58 

Substituting now for -M. its value, 
dx 

dx 2 cfiff a 2 y 3 

By differentiating again we may obtain 
d?y = 3b 6 x 

dx 3 «y 

The first differentiation may be conveniently performed by 
differentials instead of differential coefficients. Thus we should 
have from the equation 

a 2 y 2 + b 2 x 2 = a 2 b 2 , 
2a 2 ydy + 2b 2 xdx = 0, 

giving -2 — _. as before. 

dx a 2 y 

In deriving — \ J, ••-, it is better to use differential co- 
dx 2 dx 3 

efficients rather than differentials. 



EXAMPLES. 



1. tf=±ax. 4/ = 2a ? d 2 y == _^ 

dx y dx 2 y 3 



2. sin(xy) = mx. 



dy _ m — y cos (xy) 
dx x cos (xy) 



3 xV= =y* t d v = , y 2 - x y l °gy = y 2 (i - logs) 

da? x 2 — xylogx x 2 (l — log?/) 

4. y 2 —2xy = a 2 . 

dy = y dV = a 2 d 3 ?/ = 3a 2 a c^v = __a_ 2 

dx ?/ — x dx 2 (y — x) 3 dx 3 (y — x)^ dy 2 ?/ 



54 DIFFERENTIAL CALCULUS. 

5. y = sm(x + y). 

dy _ cos (# + ?/) d 2 y — y 

dx 1 — cos {x + y) dx 2 [1 — cos(x + ?/)] 3 

6. e*+"=™. **- yf"- 1 ) , ^- 2/r(^-l) 2 +(?/-l) 2 ] 

da; %(y — 1)' dar 2 x*(y — I) 3 

~ dy tan a? d 2 ?/ tan 2 ?/ — tan 2 a; 

7. seccccos2/ = m. -£ = , — ^ = — 

dx tan?/ dar tan 3 ?/ 

8. *+f-3axy = 0. *JL = _tpM % g = _ -|*SL. 

da; y z —ax dar (y'—axy 

9. x= a — b cos 0, y = aO-\-b sin 0, 

the variables being #, y, and 0. 

dy _ q + ftcosfl d*y__ b + acos 
dx bsmO ' dec 2 6 2 sin 3 



CHAPTER VII. 

EXPANSION OF FUNCTIONS. 

32. The student is probably already familiar with, methods 
of expanding certain functions into series. Thus, by ordinary 
division, 

= 1 — x + x 2 — x 3 -j : 

1 + x 

by the Binomial Theorem, 

(a + x) n = a n + na^x + n ( n - r > a «-'<x*+ •••• 

But these methods are limited in their application to certain 
forms of functions. We are now about to consider a method 
of expansion applicable to all functions, and including as 
special cases the expansions just referred to. 

These methods are known as Taylor's Theorem and Mac- 
laurin's Theorem. These two theorems are so connected that 
either may be regarded as involving the other. We shall first 
consider Maclaurin's Theorem as the simpler in expression 
and derivation. 



33. Maclaurin's Theorem. This is a theorem by which any 
function of x may be expanded into a series of terms arranged 
according to the ascending integral powers of x. It may be 
expressed as follows : 

/(^)=/(0)+/'(0)^+r(0).| 2 +r'(0)^+- 

1 l£ l£ 

in which f(x) is the given function to be expanded, and f'(x), 
f"(x), /'"(#), .••, its successive differential coefficients. 



56 DIFFERENTIAL CALCULUS. 



That is, f\x) = j-f(x), 

ax 



/"(*)=!/'(*), 



J v } dx J K J 



/(O), /'(O), /"(O), • ••, as the notation implies, denote the 
values o£ f (x) , f (x) , f" (x) , • ••, when x = 0. 

34. Derivation of Maclaurin's Theorem. This may be de- 
rived by the method of Indeterminate Coefficients by assuming 

f(x) = A + Bx+Cx?+Dx 3 +Ex*+ (1) 

where A, B, C, • • • are supposed to be constant coefficients. 

Differentiating successively, and using the notation just de- 
fined, we have 

f'(x) = B + 2Cx + 3Dx 2 +4Ex s + (2) 

f"(x) = 2C + 2.3Dx + 3AEx 2 + (3) 

f"(x) = 2.3D + 2.3.±Ex+ . (4) 

/*(a>)=2.3.4JE+ (5) 

Now since equation (1), and consequently (2), (3), ••• are 
supposed true for all values of x, they will be true when x = 0. 
Substituting zero for x in these equations, we have 

from(l), f(0) = A, or A =/(0), 

« (2), /'(0) = S, or B=f'(0), 

« (3), /»(<>) = 2 (7, or 0=£J*P, 



EXPANSION OF FUNCTIONS. 57 

from (4), /"'(0) = 2-3 D, or D =-Q91, 
" (5), / iv (0) = 2-3.4J£, or E =G21. 

Substituting these values of A, B, C, ••• in (1), we have 

x 2 
\1 



/(*)=/(0)+/'(0)|+/"(0)|+/»'(0)|+ (6) 



35. As an example in the application of Maclaurin's Theo- 
rem, let it be required to expand log (1 -f- x) into a series. 

f(x) = log (1 + x), /(0) = logl = 0. 

f'(x) = -±-=(l+x)-\ /'(0) = 1. 
1 -\-x 

f<<(x) = -{l + x)- 2 , /"(0) = -l. 

f"'(x) = 2(l + x)- 3 , /'"(0) = 2. 

/v (a . } = _ [3(1 + i)-*, /*(0) = - [3. 

jr(a0^jk(l + *)-*, / v (0) = [4. 



Substituting in (6) Art. 34, we have 

5V; 2 [3 [4 16 

or log (l + aj) = a? -- + --- + - . 

5vy 23 45 



36. If, in the application of Maclaurin's Theorem to a given 
function, any of the quantities /(0), /'(0), /"(0), ••• are 
infinite, this function is not capable of being expanded in the 
proposed series. This is the case with logx, x*, cot a;. 



58 DIFFERENTIAL CALCULUS. 

EXAMPLES. 

Derive the following by Maclaurin's Theorem : 



1/\vl /ytO /yti 
. smaj = aj — — -f-rrr 1" 



li is n 

. cosx = l— , K • 

H Li Li 

/y»6 /ytO /yi4 

3. ( r = i +a+ _ + _ + _ + .... 

4. (a + x)"= a"+ na n ' x x + n ( n ~ 1 ) a —^ 



+ n(n-l)(n-2) an _ v+ __ 



/ x 2 X 3 X* \ 

5. log a (l -1- a) = Jf ( a; — — + -- — +••• J, where Jf = log a e. 

6. tana; = aj + - + -Tr + -i 

^t. tan- 1 a: = a;--+--- + .... 
Here f(x) = tan -1 a;, 

1 +ar 
/''(z) = -2z + 4af 3 -6ar 5 +...,, 



. , r 1 or 3 , 1-3 a 5 , 1.3-5 x 7 . 

8. Bm-i* = * + -.- + — •- + — - ? + 



EXPANSION OF FUNCTIONS. 59 

Here f(x) = sm~ 1 x 1 

VI — ar 
Expanding by the Binomial Theorem, 

^ w ^2 2-4 24-6 

= 1 + aa 2 + 6x 4 +cx 6 + •••, 

, 1,1-3 1-3-5 

where a = - ? & = — , c = — , ..., 



O r 3 

9. e'seca^l + a + a^-f- — H . 

o 

10. log 10 cosa; = - m(^~ + ^ + ^ + -\ where Jf =.4342945. 

3/ £C £C £C 

11. log (l + sm^ = »--+- -- + -.-. 

12. Prom Ex. 7 derive 

? =1 _M_M_.... 

4 3 5 7 9 

Also, since tan _1 l = tan -1 - + tan -1 -, 



7T = 1_1/1 

4 2 3V2 



1(1)' 



+H(I)+I(I)-^- 

= .4636476 . • • + .3217506 . • • = .785398 • • 
tt = 3.141592-... 



80 DIFFERENTIAL CALCULUS. 

The computation includes 10 terms of the first series and 
7 of the second. 



) 



-^ 13. From Ex. 3 show that 

av-r ^ x 2 , x* . / — t/ X s . x 5 

= coscc + V— lsinx, by Exs. 1, 2. 

Similarly, show that 

e -w=l_ cos x _ - x /_i s i n Xt 

Erom these two equations derive the exponential values of 
the sine and cosine, 

since = — , 

2V^i 

cos x = 



37. Taylofs Theorem. This is a theorem for expanding 
any function of the sum of two quantities in a series arranged 
according to the powers of one of these quantities. 

As the Binomial Theorem expands (x + h) n in a series 
arranged according to the powers of h, so Taylor's Theorem 
expands any function of (x -f h) in a similar series. It may 
be expressed as follows : 

f(x + h) =f(x) +f(x)h +/"(*)£ +/"'(»)£+ .... 

38. The proof of Taylor's Theorem depends upon the fol- 
lowing principle : 

If we differentiate f(x + h) with reference to x, regarding h 
constant, the result is the same as if we differentiate it with 
reference to h, regarding x constant. 



EXPANSION OF FUNCTIONS. 61 

That is, A/(aj + h) = A/ (aj + A), 
ax a/i 

For, let 2 = x -f ft, 

then by (1) Art. 22, 

—f(x + 70 = —f(z) = ~f(z) — , 
dx Jy J dx JK ) dz JK } dx 

±f( X + h) = ±f(z) = y- A*)^' 

d7i d/i dz dli 



But 


* = 1, and ^ = 1; 
dx dli 


erefore 


s«-+*)-|a.+*). 



39. Derivation of Taylor's Theorem, With the aid of the 
preceding article we can now derive Taylor's Theorem by the 
method of Indeterminate Coefficients. Assume 

f(x + h)=A+Bh+Ch 2 +Dh s + .~ ... (1) 

where A, B, C, ••• are supposed to be functions of x but not 
of h. 

Differentiating (1), first with reference to x, then with refer- 
ence to h, 

d ~, . , N dA . dB, . dC 19 , dD 1% , 

— f(x + h) = 1 7i-\ h 2 H h 3 H , 

dx dx dx dx dx 

— f(x + h)=B + 2Ch + 3Dh 2 + .... 

By Art. 38, the first members of these two equations are 
equal to each other, therefore 

§A + d^k + <Wtf + ... = B + 2C7i + 3D7i 2 + .... 
do; da; da; 



62 DIFFERENTIAL CALCULUS. 

Equating the coefficients of like powers of h according to the 
principle of Indeterminate Coefficients, we have 

dx dx 

dB _ 2p ~, _ld 2 A 



dx 2 dx 2 

dx [3 dx 3 



The coefficient A may be found from (1) by putting h = 0, 
as the equation must hold for this value among others. 



Then 


A=f(x). 


Hence 


B = —=f'(x). 

dx J v ' 




2 da? 2 J W 




\3dx* \S J K } 



Substituting these expressions for A, B, C, ••• in (1), we 
have 

Ax + h)=f(x)+f\x)h+f%x)¥+f'%x)^ + .... . (2) 

40. Maclaurin's Theorem may be obtained from Taylor's 
Theorem by substituting x = 0. We then have 

f(h) =/(0) +f(0)h +/"(0) f +/"'(<>) f + .... 

This is Maclaurin's Theorem expressed in terms of h in- 
stead of x. 



EXPANSION OF FUNCTIONS. 63 

41. As an example in the application of Taylor's Theorem, 
let it be required to expand sin (a; -f- h) into a series. 

f(x -f- li) = sin(cc -f- h), 
f(x) = since, 
f'(x) = COS#, 
f"(x) = — sin a;, 
/'"(#) = — cos cc, 
f iv (x) = since. 

Substituting these expressions in (2) Art. 39, we find 

h 2 . h 2, W • 
sin (cc -f- A) = since -t-Acoscc — — sina; — — coscc-j- — -smoH . 

l£ L£ Li 

EXAMPLES. 

Derive the following by Taylor's Theorem : 
t i / , j,n i , & A 2 7* 3 7* 4 L 

2. (a? + ft)" = tf» + waT" 1 /i + M^" 1 ) aJ- 2 A 2 

[ n(n-l)(n-2) xn _ 3h3 ( ^ 
7* 2 , 7i 3 

4. tan (x + 7i) = tan x + li sec 2 a? + 7i 2 sec 2 cc tan x 

+ - sec 2 a(l + 3tanV) + .... 

o 

5 5. e* + »=e !e fl + 7i + — + — + .. 

\, [2 ^[3^ 

-, 6. logsin(#-f 7i) = logsin#-j-7icotcc -cosec 2 #-f~- —• 



^ 3. cos(#H-7i) = coscc — 7t since — —cos cc + — since -f- 



64 DIFFERENTIAL CALCULUS. 



7 9 

7. log sec (x -f h) = log sec x -j- 7i tan # + — sec 2 # 



7,3 7,4 

-f — sec 2 ajtanaj + — sec 2 z(l + 3 tan 2 a?) + .... 



42. The preceding proofs of Taylor's and Maclaurin's 
Theorems by the method of Indeterminate Coefficients are not 
altogether satisfactory, inasmuch as the possibility of develop- 
ment in the proposed form is assumed. 

Any rigorous proof of Taylor's Theorem, independent of 
Indeterminate Coefficients, is comparatively difficult. We give 
the following as presenting the least difficulties to the student. 

43. Continuous Functions. A function is said to be con- 
tinuous between certain values of the independent variable, 
when it changes gradually while the variable passes from one 
value to the other. In other words, a continuous function is 
one that can be represented by a continuous curve. 

44. If a given function <£ (x) is zero when x = a and when 
x = b, and is finite and continuous between those values, as 

well as its differential co- 
efficient <£'(#) ; then <f>'(x) 
must be zero for some 
value of x between a and b. 
Let the function be rep- 
resented by the curve 
y = cf> (x) . Let OA = a, 
OB = b. Then according 
to the hypothesis, y = 
when x=a, and when x = b. 
Since the curve is continuous between A and B, there must 
be some point P between them, where the tangent is parallel 
to OX, and consequently <£'(#) = 0. (See Art. 94.) Hence the 
proposition is established. 




EXPANSION OF FUNCTIONS. 65 

With the aid of this proposition Taylor's Theorem can now 
be derived without the use of Indeterminate Coefficients. 

45. Proof of Taylor's Theorem. Suppose f(x) and its suc- 
cessive n + 1 differential coefficients to be finite and continuous 
between x— a and x — a + h. Let 

+(x)=f(a+x) -f(g)-xf(a) - g/"(a) - - *>»(«) - jglg, 

where 

B = l==[/(a + h) _/ (cl )-A/'(a)-|/'' (a)-— ~fW\ 

It is to be noticed that R is independent of x. 

It is evident that cf> (x) = when x = and when x — Ti. 
Hence by Art. 44, <£'(#) = for some value of x between 
and h. Suppose h' this value. Then 



x 2 



X" 



$>( x )=fi(a + x)-f\a)-xf%a)-?-f»'(a)..--^-—r(a) 

l~ \n — l 

x n 
— — i2 = 0, when # = ft'. 



But cf>'(x) = when x— 0; hence cf>"(x) = for some value 
of x between and ft'. 

Continuing this process to n -f 1 differentiations, we find 

<F+\x) =f n+1 (a + x)-R = 

for some value of x between and ft. Let this value of x 
be Oh, where 6 < 1. 

Then f n+1 (a + 6h) = R. 

Equating this value of R with that given above, we have 



» + l 



66 DIFFERENTIAL CALCULUS. 

We may now substitute x for a, since a may have any value, 
and we have 

f(x + h) =/(*) +hf< (x) + |> (x) ... + £/»(») 

7,n+l 



46. Remainder in Taylor's Theorem. The last term 

h n+l 



\n + 



-r+*(x + 6h) 



is called the remainder after n + 1 terms. When the form of 
the function f(x) is such that by taking n sufficiently large, 
this remainder can be made indefinitely small, then Taylor's 
Theorem gives a convergent series. 

47. Failure of Taylor's Theorem. When f(x) or any of its 
successive differential coefficients are infinite or discontinuous 
between x and x + 7i, the preceding demonstration no longer 
holds good, and for such a function Taylor's Theorem is said 
to fail. 

48. Remainder in Maclaurin's Theorem. If we let x = in 
the preceding equation, we have 

/W=/(0) + V'(0) + |/"(0) - +^/ n ( Q )+^/" +1 W- 

Or, substituting x for h, 

/(*)=/(0) + */'(0)+|/"(0) - +|/"(°)+j~\/" +I ( te )- 

When the remainder, ■ ^f n+l (0x), by taking n sufficiently 

[w + 1 

large, can be made indefinitely small, the series is convergent. 



EXPANSION OF FUNCTIONS. 67 

49. Remainder in certain series. Let us apply the general 



«+i 



expression for the remainder, ■ f n+l (0x), to the develop- 
ment of e x . Here \ n 



R = 



■x 



n+l 



n + 1 



,0x 



The fraction • can be made as small as we please by 

taking n sufficiently large, whatever may be the value of x. 
Moreover, e 6x is finite ; hence R approaches zero. 
Hence the series 

T ^ |3 
is convergent for all values of x. 

It is evident that , f n+1 (0x) will have zero for its limit, 

\n±l J v ; 

whenever f(x) is of such a form that all of its successive dif- 
ferential coefficients are finite. This is the case with since and 
cos x. Hence these expansions 

x 3 , x 5 
smaj = x h — — •••» 

0OB« = l-| + g-..., 

are convergent for all values of x. 

If /(aj) = log(l + x), then the remainder is 

x n + i (-iy\n 

\n + l(l + ex) n+1 ' 

This may be expressed as 



B-i=l£ 



n + l \l + 0x 



n+l 



If x is positive and equal to, or less than, unity, R has a 
limit of zero. 



68 DIFFERENTIAL CALCULUS. 

Hence the expansion 

log(l + B ) = »_| + |_J + ... 

is convergent for positive values of x, when x = 1 or x < 1, but 
divergent, when x > 1. 



CHAPTER VIII. 

INDETERMINATE FORMS. 

50. The value of a fraction is, in general, the value of the 
numerator divided by that of the denominator. When, how- 
ever, the numerator and denominator being variable have, one 
or both, the value zero or infinity, the above definition is no 
longer applicable, and must be amended or enlarged. 

The expression, value of the fraction, must be understood to 
mean, under these circumstances, that value which the fraction 
approaches as its limit, ivhen the numerator and denominator 
approach the assigned values. We shall use it in this sense in 
the present chapter. 

It is to be noticed that this new definition of the value of 
a fraction is not necessarily confined to the cases mentioned 
above, where the ordinary definition fails, but is of general 
application, since any value of a variable fraction may be 
regarded as a limiting value. 

51. A fraction ma}* take either of the three forms, -, -, -, 

a 

(where a is a finite quantity), according as the numerator or 
denominator becomes zero, or both become zero. 

In the first case, - = ; that is, if the numerator approach 
a 

zero, and the denominator a finite quantity, the fraction ap- 
proaches zero as its limit. 

In the second case, - = oo ; that is, when the numerator ap- 
proaches a finite quantity, and the denominator zero, the frac- 
tion is increasing beyond any finite limit. 

In the third case, - is called indeterminate, for the reason that 

when both numerator and denominator approach zero, this 
alone is not sufficient to determine the limit of the fraction, 
which can only be found from the general form of the fraction. 



70 DIFFERENTIAL CALCULUS. 

For instance, consider the fraction * — — 

When x = 2, the fraction takes the form - = 0. 

3 

When x = — 1, the fraction takes the form - = oo. 

When x = 1, the fraction takes the form -, which is indeter- 
minate. 

52. To evaluate a fraction that takes the indeterminate form — 

Frequently an algebraic transformation in the given fraction 
will determine the value. If the fraction in the preceding 
article be reduced to lower terms, its value, which was before 

indeterminate when x = 1, will be found to be — 



As another illustration, consider the fraction 



2 

x-2 



Va-1-1 

When x == 2, this takes the form -. But "by rationalizing the 

denominator, we transform the fraction into 



(S-2XVS-1 +D =V ^- 1 + 1> 

X — ■" —i 

which becomes 2, when x = 2. 

53. The Differential Calculus furnishes the following method 
applicable to all cases. 

/Substitute for the numerator and denominator, respectively, 
their differential coefficients. The value of this new fraction for 
the assigned value of x will be the value required. 

To prove this, suppose the fraction 5A^i = when x = a ; 

ij/\x) 

that is, <j>(a) = 0, and ip(a) = 0. 

By Art. 50, the required value of the fraction is the limit of 

*^ C l \ as h approaches zero. 
t(a + hy . 



INDETERMINATE FORMS. 71 

By Taylor's Theorem, 

Substituting a for a;, and remembering that <£(a) = 0, 
^(a) = 0, we have 

+'(a) + <£"(«) ,| + +'"(a) .|+- 
^(a + ft) = i 14 . . . (1) 

therefore, as 7i approaches zero, 

the limit of i(M^) = £M. 
ij/(a + 7i) if/' (a) 

If <£'(a) = 0, and «//(a) = 0, we have similarly from (1), 
as h approaches zero, 

the limit of »(* + *) =£^0. 

i//(a-f-ft) V"0) 

that is, the process must be repeated, and as often as may 
be necessary to obtain a result which is not indeterminate. 
For example, let us find the value of the fraction in Art. 51, 

^ v ' = ! — = -, when x = 1. 

xp(x) x 2 ~l 

Hence ^M = ^-3 = __l when* = l. 

xp\x) 2x 2 

For another example, let us find the value of 

<f>(x) e*+e- x — 2 , A 

^ v y = — ■ = -, when a; = 0. 

if/(x) 1 — cos a; 

y , = = -, when x == 0. 

^ (#) sin a; 

ffi = «l±£L* = 2 , when a- = 0. 
tp"(x) cos a? 



72 DIFFERENTIAL CALCULUS. 

EXAMPLES. 

Find the values of the following fractions : 
, Iqq- x 



2. 
3. 

4. 



x- 1 
x-2 

(x-iy-i 

e x — e~ x 



since 
x sin x 



2 since 



b lo g(2^-l) ? 
tan (x — 1) 



x=l. 


Ans. 1, 


x = 2. 


Ans. -, 


x = 0. 


^.?2S. 2 


x = 0. 


J.71S. 


x=l. 


^Iws. 4, 



6. tau ^~^ , <c = 0. ^5. 2. 



x — sin £ 



- log sin x 

(7r-2cc) 2 ' 2 



7T 

# = -• 



e x_ e -*_2a; „ 

8. — ■— , #=0. 

x — since 



9. , x=l. 

x 6 -lbx 2 + 2±x-10 

") 1Qb 2tans-sin2s ^ = Q> 



e Ss_ 10e 2z+3 + 15e x+4_ 6e 5 

11. ' , X=l. 

e 4x_ 6e 2x+2 + 8e x + 3_o e 4 

10 sec 2 # — 2 tan cc tt 

12. , x = ~r 

1 + cos4ce 4 

7 13. _i£±^!_, x=2. 

(a; — ±)e x + e 2 x 





1 




8 


Ans 


. 2 


Ans. 


1 




10 


Ans, 


. 2, 


Ans. 


5e 




2 


Ans. 


1 




2 


Ans. 1 


3e 4 . 



INDETERMINATE FORMS. 73 



54. A fraction may take either of the forms, — , — , —. 

a oo go 

By regarding the value of a fraction as a limit, it is evident 

that in the first two cases, — = oo, and — = 0. 

a oo 

The form — is indeterminate, for the reason that, if the 

00 

numerator and denominator both increase beyond any finite 
limit, this alone is not sufficient to determine the limit of the 
fraction. 

00 



CO 



55. To evaluate a fraction that takes the form 

Suppose £i_ J = _ } when x = a ; 
i[/(x) GO 

that is, <K a ) == °°) an( i <K a ) = °°- 

By taking the reciprocals of <£(#) an( i <A0*0> we nave 

when x = a. 







*(*) 


^(x) 









*(*) 


1 


~o' 


Hence 


by 


Art. 53, 







the limiting value of ilL-l, when x = a, is the value of 






dx \xf; (x) J _ [if; (X) J _ i}/'(x) 



df 1 \ $'(x) ^(s) 



L<K*0. 



, when a? = a. 



That is, 



<f>(a) ==: ^(a) 
iff (a) <£'(a) 



L^(a)J' 



(1) 



hence 1== «AW^M or iio) = *W. . . (2) 

+'(0)1,(0)' 4, (a) +'(a) y ) 



"¥75/1 



74 DIFFERENTIAL CALCULUS. 

In deriving (2), we have divided (1) by ^W . If, however, 

6(a) ^ (a) 

^ ( =0 or oo, equation (2) does not logically follow from 

xp(a) 

(1). Nevertheless, it may be shown that (2) is true in these 
cases also. 

Suppose £L_Z = 0, and n a finite quantity, 

then ^M + n = iM±MM = n . 

To this last fraction, (2) evidently applies, 

therefore <K«) + niK«) = *'(<*) + ^'(a) 

^r(o) • f (a) 

thatis, W+^^M + n, or 4^1 = ^ 
<K«) <A» *(a) f(a) 

If ^M = *o, then £^>=0, 

and we have the preceding case. 

Thus the form — is evaluated in the same way as the form — 
J GO 

For example, find the value of 



° , when x = 0. 
cot a; 

Here *W = ^ = «, 

j//(^) COt£ 00 
1 

<f>'(x) = 

ij/' (x) — cosec 2 x" 

<£"(#) _ 2sin#cosa; _0 
<^" O) ~ 1 

56. To evaluate a function that takes the form 0-co. 
The product <f>(x)'\j/(x) becomes indeterminate when one 
factor = 0, and the other = oo. 







when 


X 


= 0. 


sin 2 a; 




: o' 


when 


X 


= 0. 


X 


4=c 




when 


X 


= 0. 



INDETERMINATE FORMS. 75 

By taking the reciprocal of one of the factors, the expression 

may be made to take the form - or — 

oo 

For example, find the value of 

(-7T — 2 x) tan x, when x = — 

u 

This takes the form -oo. But 

(7T — 2x) tancc= - = -, when « = - 

v } cotx 2 

The value is found by Art. 53 to be 2. 

57. To evaluate a function that takes the form oo — oo. 
Transform the expression into a fraction, which will assume 
either the form - or — 

00 

For example, find the value of 

, when x=l. 

log x x — 1 

This takes the form oo — oo. But 

1 1 x — 1 — loo; x , - 

= -, when x= 1. 



logo? x—1 (x— l)loga? 

The value is found by Art. 53 to be - 

J 2 

EXAMPLES. 
Find the values of the following : 

loo- X — 



1 


\ z / 


when 


7T 

x = — 
2 


A71S. 


0. 




tana; ' 


2. 


sec 3 a; cos 7 #, 




x = *. 
2 


Ans. 


7 
3 


3. 


sec a; — tana?, 




* = =. 
9 


Ans. 


0, 



76 DIFFERENTIAL CALCULUS, 



4. 


i 

(a s — 1)#, 


#=00. 


Ans. log a. 


5. 


log cot X 
cosec x 


#=0. 


J.71S. 0. 


6. 


1 

cosec" x -, 

X 2 


x = 0. 


^.71 S. — 

3 


7. 


e 1 


x=l. 


2 


e x — e x — 1 


8. 


(1 — tana?) sec 2 a?, 

sec — 


4 


J.ns. 1. 


9. 


2 


<c=l. 


-4?iS. 00. 




log (1-aj)' 




10. 


(a 2 — ce 2 ) tan — , 


x = a. 


Ans. *£ 




2a 




7T 


11. 


log tan 2 x 
log tan cc ' 


IT 

x = — 

2 


^IftS. — 1. 


12. 


2 1 


x=0. 


Ans. — 

2 


sin 2 a? 1 — cos a;' 


13. 


2 x t an a? — 7r sec x, 
tang (*+!)] 


7T 
fl? = — 

2 


.^ns. —2. 


14. 


a?=l. 


J.ws. 2. 




, 7ra; 
tan — 

2 







1 / 7rX . , 7ra?\ 

log ( sec— + tan— ) 

15. ^ i £Z, z = l. ^rcs. -1. 

log 0-1) 

58. To evaluate a function that takes either of the forms, 
0°, oo°, I 00 . 

Take the logarithm of the given function, which will assume 
the form 0-oo, and can be evaluated by Art. 56. From this 
the value of the given function can be found. 



INDETERMINATE FORMS. 77 

For example, find the value of 

i 
(l-\-x) x when x = 0. 
This takes the form 1 °°. 

Let y = (l+x) x ; 

then \ogy = - log (1 -\- x) =co-0, when x = 0. 

The value of log y is found to be 1. Hence the value of 
y is e. 



EXAMPLES. 



Find the value of the following 


: 




i. (i + x*y, 




when 


x = 0. 


Ans. 1. 


2. (e*+l)% 






X= 00. 


^l?is. e. 


i 
3. (cos 2 a;) *\ % 






35=0, 


Ans. — 

e 2 


4. x^, 


\ 




05=1. 


Ans. -. 
e 


5. (logic)* -1 , 






X=l. 


Ans. 1. 


•• H)' 






X = CC. 


-4ws. e a . 


7. (cota) sin % 






x = 0. 


Ans. 1. 


8. (sina;) tana: , 






7T 

x= -• 

2 


-4ns. 1. 


a 






x = l. 




9. (oj— l) logsin7rx , 


^Lns. e a . 


.0. (tenfrf, 






x=l. 


^4ns. -. 



78 


DIFFERENTIAL 


CALCULUS. 




11. 


tan — , 


a? = l. 


Ans. 1. 


12. 


( 2 -«J "' 


# = a. 


2 


13. 


1 

(cota) logx , 


£C = 0. 


^4ns. — . 
e 


14. 


1 
[}og(e + x)f, 


x = 0. 


i 


15. 


(logxy, 


x=0. 


-4ras. 1. 


16. 


i 

(e* + x)% 


x=0. 


Ans. e 2 . 



CHAPTER IX. 

PARTIAL DIFFERENTIATION. 

59. Functions of several Independent Variables. In the pre- 
ceding chapters differentiation has been applied only to func- 
tions of a single independent variable. We shall now consider 
functions of two or more independent variables. 

60. Partial Differential Coefficients. Representing by u a 
function of the two independent variables x and y, 

u=f(x,y) (a) 

If we differentiate (a), supposing x to vary and y to remain 

constant, we obtain — 
dx 

If we differentiate (a), supposing y to vary and x to remain 

constant, we obtain — 
dy 

The differential coefficients, — , — , thus derived, are called 

dx dy - a 

■* ou ou 

partial differential coefficients and are denoted by t - ? t~' 

For example, if u = X s + 3 x 2 y — y s , 

~ = 3x 2 + 6xy, regarding y as constant. 
dx 

— — 3x 2 — 3y 2 , regarding x as constant. 
dy 

In general, whatever the number of independent variables, 
the partial differential coefficients are obtained by supposing 
only one to vary at a time. 

79 



80 DIFFERENTIAL CALCULUS. 



EXAMPLES. 

1. If u = x 3 f-2xy* + 3x 2 tf, 

show that x— + y— = 5u. 

dx dy 

2 . . = („-*)(*-*)(*-„), £+g+£-a 

3. .-]og<? + , + '-S** £+£ + £ = _*_ 

4. % = , — - + — = (x + 2/ — 1) u. 

e x + e y dx dy- * J 



du , du 



5. u = log (a? + Var 2 + 2/ 2 ), x ^~ + 2/"=r = 1 



dx 3y. 



6. u = e x sin y + e y sin a?, 



'^Y+ (yf= e2x + e 2y + 2 e*+* sin (x + y). 

7. w = log (tan a? + tan 2/ + tans;), 

• o du . . du . ■ o du o 

sin 2 a?— + sin 2?/ h sin 2s; — = 2. 

ox dy dz 

61. Pai'tial Differential Coefficients of Higher Orders. By- 
successive differentiation, regarding the independent variables 
as varying only one at a time, we may obtain 

d 2 u d 2 u d 3 u d A u 
dx 2 ' dy 2 ' daf*' dy 4 ' 

If we differentiate u with respect to x, then this result with 

respect to y, we obtain — ( — ], which is written 

dy\dxj dydx 

Similarly, — — is the result of three successive differentia- 
tor 
tions, two with respect to x, and one with respect to y. It will 
now be shown that this result is independent of the order of 
these differentiations. 



J 



That is, 



PARTIAL DIFFERENTIATION. 81 

d 2 u d 2 u d 5 u d 3 u d 3 u 



dydx dxdy dydx 2 dxdydx dx 2 dy 

62. Given u =f(x, y) (a) 

+i 4. d fdu\ d fdu\ 

to prove that — — - = —( — ]• 

dy\dxj ax\dy J 

Supposing x alone to change in (a), 

Am = f(x + Ax, y)-f(x, y) . . . . / & x 

Ax Ax W 

Now supposing ?/ alone to change in (b), 
_A/Au\ _ f(x+Ax, y+Ay)-f(x, y+Ay)-f(x+Ax, y)+f(x, y) 



Ay\AxJ Ay Ax 

Eeversing the above order, we find 

Au = f(x, y + Ay) -f(x, y) 
Ay Ay 



A_/A%\ f(x+Ax, y+Ay)-f(x+Ax, y)-f(x, y + Ay)+f(x, y) 



and 

-)=■ 

Ax\AyJ Ax Ay 

Hence A(AA = AfA\ 

Ay\AxJ Ax\AyJ 

This being true, however small Ax and Ay may be, we have 

for the limits of the above 

d fdu\ _ d fdu\ d 2 u _ d 2 u 

dy\dxj dx\dyj dydx dxdy 

63. This principle, that the order of differentiation is imma- 
terial, may be extended to any number of differentiations. 
Th d 3 ^ __ d 2 /du\ _ d 2 fdu\ _ d 3 u 



dydx 2 dydx\dxj dxdy\dxj dxdydx 

= d f d " H \ = d ( dhl \ _ dht 
dx\dydxj dx\dxdyj dx' 2 dy 

It is evident that the principle applies also to functions of 
three or more variables. 



J* 



82 DIFFERENTIAL CALCULUS. 



EXAMPLES. 

d 2 u d 2 u 

Verify = , in each, of the four following equations : 

oydx , oxoy 

1. u = y log (1 + xy). 3. u == sin (xy 2 ). 

o v A ax— by 

2. u = £C y . 4. w = ^-« 

a?/ — &ic 

T 1 II f/ - ?/ fill fi'if 

5. If w = - — — , show that #— — + y = 2 — 
a; + y dar dxay da? 

6. « = («• + /)*, 3z^ + 3*/^ + ^ = 0. 

dxay oy £ oy 

7. u = *~, j^-=(l+3xyz + xy-z*)u. 

oxoyoz 

x y z 8^U - - - 

8. % = 2/%a + 2 Ve* + afy^ 2 2 2 = e 2-f e 2 + e2. 

9. w = sin (y + 2) sin (z + a;) sin (x + ?/), 

j^-==2cos(2x + 2y + 2z). 
oxoyoz 

64. TotaZ Differential. If w is a function of two or more 
variables, and all vary at the same time, the change in u is 
called the total increment, and if infinitely small, the total dif- 
ferential of u. 

This total differential of u may be obtained by the usual 
formulae of differentiation, using differentials as in Art. 28. 

Tor example, suppose 

u = a?y — 3 x 2 y 2 . 
Differentiating, regarding both x and y variable, 
du = d(x 3 y)-d{3x 2 y*) 

= a?dy + yd(x s ) - 3 M(f) - 3 yHif) 
= My + 3 a^ycZo? — 6 a%cfa/ — 6 xy 2 dx 
= (3 a; 2 ?/ — 6 xy*)dx + (jb 3 — 6 x 2 y)dy. 



TOTAL DIFFERENTIAL. 83 

But 3a?y-6xy 2 = — , and ar 3 -6afy = — • 

dx dy 

Hence du = — dx-\ dy (1) 

dx dy w 

This expression for the total differential holds for any func- 
tion of two variables, u=f(x, y). 
■■'/ For, if we differentiate this equation, using differentials as 
in the preceding example, we may arrange the terms in two 
groups containing dx and dy respectively, so that the result 

will be of the form 

du = Pdx -f Qdy (2) 

Now if x alone varies, y being constant, (2) becomes 

d x u = Pdx, giving — = P. 
dx 

If y alone varies, x being constant, (2) becomes 

d y u = Qdy, giving -^ = Q. 
dy 

Substituting in (2) these expressions for P and Q, we have 

du = — dx-\ dy (3) 

dx dy u w 

Similarly, if u =/(#, y, z), it may be shown that 

du = -^-dx + ^dy + -^dz . . . . (4) 
dx dy dz 

65. The result of the preceding article may be reached also 
in the following way. The toted differential of a Junction of 
several independent variables is the sum of its partial differen- 
tials arising from the separate variation of the variables. 

Let Au, du, denote the total increment, and differential of u. 

A x u, A y u, d x u, d y u, the partial increments and differentials, 
when x and y vary separately. 

Let u =f(x, y), 

u' =f(x + Ax, y), 

u" =f(x-\- Ax, ?/+ Ay). 



84 DIFFERENTIAL CALCULUS. 

Then A x u = u' — u, 

A y u' = u" — u', 
Au = u" — u. 
Hence Au = A x u + A y iiK 

Now if A#, Ay, and consequently A x w, A y %', Au, become infi- 
nitely small, we have 

du == d x u + d y u, 
since the limit of u' is u. 

We may write d x u = — dx, du = — d?/, 
dx dy 

giving du — — dx -\ dy. 

dx dy 

The process above may be extended to functions of three or 
more variables. 

EXAM PLES. 

Find as in Art. 64 the total differential of u in each of the 
following, and show that it agrees with (3), Art. 64. 

1 . u = ax 2 + 2 bxy -f cy 2 , du = 2 (ax + by) dx-\-2 (fix + cy) dy. 

2. u = x^», du = u ( 1 Ml dx + ]2M d y\ 

V x y ) 

3. u — log ^ + 2 tan -1 -, du = — (ydx — xdy). 

x + y y x*-y 4 

Find the total differential of u in each of the following, and 
show that it agrees with (4), Art. 64. 

4. u = ax 2 + by 2 + cz 2 + 2fyz + 2gzx + 2 hxy, 

du = 2 (ax -f- hy + gz) dx-\-2 Qix + by +fz)dy + 2 (gx +fy + cz) dz, 

5. u = x yz , du = x yz ~ Y (yzdx -f- zx log xdy + £?/ log xdz). 

6. it = tan 2 x tan 2 # tan 2 z. du=±u( '-J^L- + _J?L + - T %-\ 

ysm 2 x sin 2 ?/ sin 2 2/ 



PARTIAL DIFFERENTIATION. 85 

66. Condition for an Exact Differential. 
The expression Pdx+ Qdy is called an exact differential, when 
it is the total differential of some function of x and y. 

For example, ydx + xdy is an exact differential, 
because it is the differential of xy. 

But 2 ydx + xdy is not an exact differential, 

because it is not the differential of any function of x and y. 

The general expression Pdx + Qdy is an exact differential 
only when P and Q satisfy a certain condition, which we will 
now derive. 

Suppose this expression to be the differential of some func- 
tion u, of x and y. 



Then 


du = Pdx + Qdy. 


But from (3) Art. 64, 


du — — dx-\ dy. 

dx dy 


Hence 


p du ^ du 

dx' dy 



Differentiating the first of these equations with respect to y, 
and the second with respect to x, we have 

aP = _5^ dQ = d 2 u t 
By dydx dx dxdy 

Hence ^=^ (1) 

dy dx 

which is the condition that Pdx + Qdy may be an exact differ- 
ential. 

Similarly, it may be shown that Pdx + Qdy -f JRdz is an 
exact differential, when 

BP = dQ dQ = dR BR = 8P (2) 

dy dx' dz dy' dx dz* 



86 DIFFERENTIAL CALCULUS. 

EXAMPLES. 

By means of (1) determine which, of the following expres- 
sions are exact differentials : 

1. (3xy + 2y 2 )dx+(x 2 + 2xy)dy. 

2. (3x 2 y + 2xf)dx+(x* + 2x 2 y)dy. 

3. (xy — y 2 + 1) dx + (x 2 — xy — 1) dy. 

4. e* y \_(xy -f + l)dx + (x 2 - xy -l)dy~\. 

Show that condition (1) is satisfied by the answers to Ex- 
amples 1, 3, Art. 65; and conditions (2) by the answers to 
Examples 4, 5, Art. 65. 

67. Differentiation of an Implicit Function. The differential 
coefficient of an implicit function may be expressed in terms 
of partial differential coefficients. 

Suppose y and x connected by the equation <£ (x, y) = 0. Let 
u represent the first member of this equation. That is, 

u = cf>(x,y) = (1) 

From (3) Art. 64, we have for the total differential of u, 

du = — dx H dy. 

dx dy 

But by (1), u is always zero, that is, a constant ; and there- 
fore its total differential du must be zero. Hence 



(2) 







du -, , du 7 A 
■— dx + — dy = 0, 
dx dy 






,\ ^ = 


du 

dx 




dx 


du 
dy 


For 


example, 


suppose, as in Art. 31, 
a 2 y 2 + tfx 2 - a 2 b 2 = 0. 



PABTIAL DIFFERENTIATION. 87 

Here u = a 2 y 2 + b 2 x 2 — a 2 b 2 , 

and — = 2b 2 x, — = 2a 2 y. 

ox dy 

Hence by (2), dy^_2_Vx^_bH 
J w dx 2a 2 y a 2 y 

Derive by (2) the expressions for -^ in the examples in 
Art. 31. dx 

68. Extension of Taylors Theorem to functions of two inde- 
pendent variables. If we apply Taylor's Theorem to 

f(x + h,y + k), 

regarding x as the only variable, we have 

fix + h,y + k) =f(x, y + k) + h^M y + k) 

Now expanding f(x, y + k), regarding y as the only variable, 
fix, y + k) = f(x, y) + Tc^-fix, y) +|^ 2 /fe y) + -. 

Substituting this in (1), 
X* + A, 2/ + &) =/& 2/) + h -fix, y) + k jif(x, y) 

This may be expressed in the symbolic form thus : 
fix + h,y + k) =fix, y) +^A + fclW 2/) 



88 DIFFERENTIAL CALCULUS. 

where ( h- — h&— ) is to be expanded by the Binomial Theorem, 

as if h — and 7c — were the two terms of the binomial, and the 
dx by 

resulting terms applied separately to f(x, y). 

69. Taylor's Theorem applied to functions of any number of 
independent variables. By a method similar to that of the pre- 
ceding article we shall find 

f(x+h, y+7c, z+T)=f(x, y, z) + (h£ + ^+ l £)A x > V' z ) 
This expansion may be extended to any number of variables. 






CHAPTER X. 

CHANGE OF THE VARIABLES IN DIFFERENTIAL 
COEFFICIENTS. 

dfj dii d 3 u dx d x d x 

70. To express -f, -J, -J, ... in terms of -, -, -, •••. 
dx dx 1 dx 6 dy dy 1 dy 6 

This is called changing the independent variable from x to y. 
t>tt n\ Avf 91 dy _ 1 / a ) 



j-»j V^/ -"-*«• «-*> 


dx dx 
dy 




By (1) Art. 22, 


d 2 y _ d dy _ d dy dy 
dx* dx dx dy dx dx 




From (a), 


^-~ d?x 

d dy d 1 dy 2 
dy dx ~ dy dx~ /dx\ 2 ' 
dy [dyj 

d 2 x 




•"• 


d 2 y dy 2 
dx 2 ~ (dx\ s 

W 


• • 


Similarly, 


d 3 y __ d d 2 y _ d d 2 y dy 
dx 3 dx dx 2 dy dx 2 dx 




From (6), 


fd 2 x\ 2 dx d 3 x 
d d 2 y \dy 2 J dy dy 3 
dy dx 2 AfccY 
\dy) 




• • 


„fd 2 x\ 2 dx d 3 x 
d 5 y \dy 2 J dy dy 3 
dx*~ fdxV> 


• • 



(b) 



(c) 



90 DIFFERENTIAL CALCULUS. 

71. It is sometimes necessary in the differential coefficients, 

dy d 2 y d 3 y 

dx* dx 2 * da?' 

to introduce a new variable z in place of x or y, z being a given 
function of the variable removed. 

There are two cases, according as z replaces y or x. 

72. First. To express -^, _Jl —M. ••• in terms of — , 

dx dx 2 dx? dx 

-—, — -, •••, where y is a given function of z. 
For example, suppose y = z 3 > 

Then d JL=^- d ±. 

dx dx 



dx 2 \clxj dx 2 

dx 3 \dxj dxdx 2 dx 3 

djii d°ii 
Similarly, __£, _ ^, •••, may be expressed in terms of z 
dx* dx° 

and x. 

It is to be noticed that in this case there is no change of the 

independent variable, which remains x. 

73. Second. To express -M, _?, _J(, ••• in terms of ^, 

da; dx 2 dx? dz 

~y, — , •••, vjhere x is a given function of z. 
dz 2 dz 3 

This is called changing the independent variable from x to z. 

For example, suppose x = z*. 

By (1) Art. 22, */ = #* 

dx dz dx 



CHANGE OF VARIABLE. 91 



But — = 3z 2 , ^- = z -l 

dz dx 3 



dx 3 dz 



<«) 



dec 2 d# dx dz dx dx 



dar 9\ dr dz/ 



. ., n d 3 w d d 2 y dz 

Similarly, _£ = _ * _. 

dar cfecfcr da? 






EXAMPLES. 

Change the independent variable from x to y in the two fol- 
lowing equations : 

1. a (**\'-*V.*»-**(&f„o. Ana. ^ + ^ = 0. 

\dx 2 J dxdx* <to?\dxj dy 3 dy 2 



(»«2«)(SH°2 + >) 



dydPy^ 
dx dx 3 






92 DIFFERENTIAL CALCULUS. 

Change the variable from y to z in the two following equa- 
tions : 

dx> 1+y 2 {dx)' * 

■ Ans. * sf^Y-ooA. 
dx l \dxj 

K J) \dx* y ) \dx) v yj dxdx* y T 

Ans. (z + l) — = ^— + z 2 +2z. 

dx? dx dx- 



Change the independent variable from x to z in the following 
equations : 

5 g + l|g + a?= ^ Ans.Af + V = 0. 

dx z xdx dz* dz 

/j d 2 y . 2x dy . y A , 

da? 2 1 -f- ar da; (1 -+- ar) 2 

d 



Ans. -^ -f- 2/ = 0. 



7. (2aj-l) 3 ^ + (2a3-l)^ = 22/, 2#=l+e*. 

(XX/ CLX 

Ans. A c il-12p{ + df-y = 0. 
dzr dz' dz 

8, a 4 ^+6a^ + 9ar . T+3a?^ + y = loga?, a;=e*. 

da? 4 dar dar da; 

. d 4 ?/ . d 2 w . 
^ S -^ 4 + 2 d? + 2/ = " 






APPLICATION TO PLANE CURVES. 



CHAPTER XL 



CERTAIN CURVES IN THE FOLLOWING CHAPTERS. 



74. We give in this chapter representations and descrip- 
tions of some of the curves used as examples in the following 
chapters. 

RECTANGULAR CO-ORDINATES. 



75. The Cissoid, 

x* 




r 



2a — x 



This curve may be constructed 
from the circle ORA (radius, 
a) by drawing any oblique 
line OM, and making 

MP = OR. 

The equation above may be 
easily obtained from this con- 
struction. The line AM par- 
allel to Y is an asymptote. 



94 



DIFFERENTIAL CALCULUS. 



76. The Witch, 




This curve may be constructed from the circle ORA 
(radius, a) by drawing any abscissa MR, and extending it 
to P by the contruction shown in the figure. 

The equation above may be derived from this construction. 
The axis of X is an asymptote. 



of 



77. The Curve, a?y = — —ax* + 2a 3 . 

o 




BEPRESENTATION OF CURVES. 



95 



78. The Catenary, 

This is the curve of a cord or 
chain suspended freely between 
two points. 




79. The Parabola referred to Tangents at the Extremities of 
the Latus Rectum, x* + y* = a?. 

OL=OL' = a. 

Y 




The line LL' is the latus rectum ; its middle point F, the 
focus ; OFM is the axis of the parabola ; the middle point of 
OF, A, is the vertex. 



96 



DIFFERED TIAL CAL C UL US. 



80. The Hypocycloid of Four Cusps, 
Y 



X s -\- y 3 = a 3 . 

This is the curve 
described by a 
point P in the 
circumference of 
the circle PR, as 
it rolls within the 
circumference of 
the fixed circle 
X, ABA}, whose ra- 
dius, a, is four 
times that of the 
former. 




The equation is that of the ellipse 



'tHf 



i. 



with the second exponent changed from 2 to f . 



REPRESENTATION OF CURVES. 

82. The Curve, a 4 y 2 = a?x 4 — x 6 . 

Y 




83. The Curve, 



a n ~ l y = x n , 



97 



(1) 



where one co-ordinate is proportional to the nth power of the 
other, is frequently called the parabola of the nth degree. 



84. If n = 3 in (1) Art. 83, we have 
The Cubical Parabola, a 2 y = x 3 . 

Y 




98 



DIFFERENTIAL CALCULUS. 



85. If n = f, in (1) Art. 83, we have 



TJie /Semi-Cubical Parabola, 
Y 



a 2 y = sc 2 , or ay 1 = ar 




POLAB CO-ORDINATES. 

86. The Circle, r = a sin 0. 

The circle is OP A (diameter, a) tangent to the initial line 
OX at the pole, 0. 




87. The Spiral of Archimedes, r = aO. 

In this curve r is proportional to 0. Assuming r = OA, 
when = 2 w, then 



REPRESENTATION OF CURVES. 



99 



The dotted part of the curve corresponds to negative values 
of 6. P. 




r — e c 



88. The Logarithmic Spiral, 

Starting from A, where 
= and r=l, r increases 
with 0; but if we sup- 
pose negative, r de- 
creases as numerically 
increases. Since r = 
only when — — oo, it 
follows that an infinite 
number of retrograde 
revolutions from A is 
required to reach the 
pole 0. 

A property of this * 
spiral is that the radii vectores OP, OP^ OP., •••, make a 
constant angle with the curve. 




100 



89. The Parabola 



DIFFERENTIAL CALCULUS. 
2 



a sec 



The initial line OX is the axis of the parabola ; the pole 
is the focus ; LL', the latus rectum. 




90. The Lemniscate, r 2 = a 2 cos 2 9. 

This is a curve of two loops like the figure eight. 

It may be defined in connection with the equilateral hyper- 
bola, as the locus of P, the foot of a perpendicular from on 
PQ, any tangent to the hyperbola. 

The loops are limited by the asymptotes of the hyperbola, 
making 

TOX =TOX= 45°. OA = a. 

The lemniscate has the following property : 

If two points, F and F', be taken on the axis, such that 

OF=OF' = —, 

V2 

then the product of the distances P'F, PF', of any point of 
the curve from these fixed points, is constant, and equal to the 
square of OF. 



REPRESENTATION OF CURVES. 



101 



The points F and F' are called the foci of the lemniscate, 
and this property may be used as a definition of the curve. 

T 



91. The Curve, 



r = a sm J 




lOf 



DIFFERENTIAL CALCULUS. 



92. The Cardioid, r = a(l — cos 0). 

This is the curve described by a point P in the circumfer- 
ence of a circle PJ. (diameter, a) as it rolls upon an equal 
fixed circle OA. 

Q 




Or it may be constructed by drawing through 0, any line OB 
in the circle OA, and producing OB to Q and Q', making 
BQ = BQ'=OA. 

The given equation follows directly from this construction. 



93. The Curve, 



r = a sin 2 0. 




CHAPTER XII. 



DIRECTION OF CURVE. TANGENT AND NORMAL. 
ASYMPTOTES. 

94. Direction of Curve. When the equation of the curve is 
given in rectangular co-ordinates, its direction at any point 
is determined by the angle made by its tangent at that point 
with the axis of X. We shall 
denote this angle by <£. 

Let P be a point in a curve 
whose equation is y=zf(x), its 
co-ordinates being x=OM, and 
y = PM. Draw the tangent 
PT, and PR parallel to OX. 
Then TPR = <$>. 

Now give to x the increment Jr 
Ax = MN-, then y will receive 
the increment Ay = QR, and we have another point Q in the 
curve. Draw PQ. 




Then 



tan QPR = 



QR 
PR 



Ay 
Ax 



(a) 



Now if Ax be supposed to diminish and approach zero, Ay 
will approach zero, the point Q will move along the curve 
towards P, and PQ will approach in direction PT as its limit. 

Taking the limits of the two members of equation (a), we 
have 

limit of QPR = TPR = <£, 

limit of — y~ — -^, 
A x dx 

dy 
dx 



and 



by definition. 



tan <£ 



a) 



104 



DIFFERENTIAL CALCULUS. 



For example, find the direction at any point of the parabola 

y 2 =4:ax. 




Here 



hence 



dy 
dx 



Va 
x' 

tan<£ =\ — 

\ x 



At the vertex 0, where x = 0, 
tan <j> = oo, <£ = 90°. 

At L, where x = a, 

tan<£ = l, <£ = 45°. 

For that part of the curve 
beyond L, as x increases, tan<£ 
and </> decrease. Thus the par- 
abola is more nearly parallel to 
OX, the further it extends from 0. 



95. Subtangent and Subnormal. Let PT be the tangent, 
Y and PN the normal, to a 

curve at the point P, whose 
ordinate is y = PM. Then 
MT is called the subtangent, 
and JOT" the subnormal, cor- 
responding to the point P. 
To find expressions for 
these quantities : 

Subtangent = MT = PM cot PTM = y cot <j> = J^- = y ~- 

dx 
dy 




Subnormal = MN= PM tan MPN~= y tan <£ = y 



dx 



that 



The length PA 7 " is sometimes called the normal. It is evident 
PN= PMsec <j> = yJl + 



dx, 



DIRECTION OF CURVE. 105 



EXAMPLES. 

x 5 

1. Tlie equation of a curve is a 2 y = aa? 2 + 2 a 3 . 

) o 

(a). Find <£ when x = and a: = a. Ans. <£ = and 135°. 
(b). Find the points where the curve is parallel to X 

Ans. x = and x = 2 a. 

(c). Find the points where <£ = 45°. Ans. x = (l± -\/2)a. 

(d) . Find the point where the direction is the same as that 

at x = da. Ans. x = — a. 

2. Where is the curve y(x — 1) (x — 2) = a; — 3 parallel to 

X? ^4ns. a = 3±V2. 

3. Show that the ellipse — + *- = 1, and the hyperbola 

18 8 

£c2= 2/ 2 -j- 5, intersect at right angles. 

4. At what angle does the circle x 2 -\-y 2 =8ax intersect the 

x 3 
cissoid y 2 — 



2a — x 
Ans. At the origin, 90° ; at the two other points, 45°. 

5. At what angle does the parabola x 2 =<lay intersect the 

witch y = o 8a3 o ? Ans. tan- 1 3 = 71° 33' 54". 

ar+4a 2 

6. Find the subtangent and subnormal of the parabola 

y 2 = 4 ax. Ans. 2x and 2 a. 

7. Find the subtangent and subnormal of the parabola of the 

nth degree y n = a^x. j_ nSm nx an( j iL. 

71X 
X 3 

8. Find the subtangent of the cissoid ?/ 2 = 




XX o 

9. Fina the normal of the catenary y = - (e a 4- e ") . Ans. -—• 



106 



DIFFERENTIAL CALCULUS. 



96. Direction of Curve. Polar Co-ordinates. By means of 
the equations 

x = r cos 0, y = r sin 0, 
we may express tan <£ in terms of r and 0. Thus 



tan <£ 



dy 
dx 



dy 
d£ 
dx 

To 



r cos + — sin $ 
dO 

— r sinfl H cos 

dO 



(a) 



The angle OPT between the tangent and the radius vector 

may also be ex- 
pressed. Denote 
this angle by \p. 
Q Let r, 0, be the 
co-ordinates of P; 
r + Ar, + A6, 
the co-ordinates 
of Q. Describe 
the arc PR about 
as a centre. 
Then 

i£Q = Ar, POR = A0, PR = rA0. 

If we suppose Q to approach P, the figure PPQ will approach 
more and more nearly a right triangle, P being the right angle. 
We have at the limit 

tan POP = — — == > 

Y i£<3 Ar 




or 



tan^ 



rdO 
dr 



r 
dr 

dO 



(*) 



We also have 
PTX 

or <f> 



OPT + POX, 



(c) 



DIRECTION OF CURVE. 



107 



97. Polar Subtangent and 
Subnormal. 

If through. 0, NT be drawn 
perpendicular to OP, OT is 
called the polar subtangent, and 
ON the polar subnormal, cor- 
responding to the point P. 

OT= OP tan OPT; that is, 

Polar subtangent = rtan^ = — • 

d0 
ON = OP cot PNO; that is, 

Polar subnormal = r cot \b = — 
r d0 




EXAMPLES. 

1. In the circle r = asin0, find if; and <f>. 

Ans. if/ = 0, and $ = 20. 

2. In the logarithmic spiral r = e a9 , show that ^ is constant. 

3. In the spiral of Archimedes, r = a0, show that tan if/ = 6 ; 

thence find the values of \p when = 2-n- and 47r. 

.4ns. 80° 57' and 85° 27'. 
Also show that the polar subnormal is constant. 

4. The equation of the lemniscate referred to a tangent at its 

centre is r^ a 2 sin 20. Find ij/, <j>, and the polar sub- 
tangent. 

Ans. if; = 26; <£ = 3 ; subtangent = a tan 2 Vsin20. 

n 

5. Given the equation of a curve r = asin 3 -; show that 

<£ = 4«/^. 3 



6. In the parabola r = a sec 2 -, show that cf> -\- ij/ = n. 



108 DIFFERENTIAL CALCULUS. 

7. In the cardioid r = a(l — cos 0), find <j>, iff, and the polar 

subtangent. 

Ans. d> = — ; »£ = -; subtangent = 2 a tan- sin 2 — 
^22 22 

8. Find the area of the circumscribed square of the preceding 

cardioid, formed by tangents inclined 45° to the axis. 

Ans. ?I(2+V3)a 2 . 

9. Derive equation (a) from equations (6) and (c), of Art. 96. 

98. Differential Coefficient of the Arc. Rectangular Co-ordi- 
nates. In the figure of Art. 94, let s denote the length of the 
arc of the curve measured from any fixed point of it. 

Then s = arc AP, As = arc PQ. 

We have sec QPR = ?Q. - M& 

PR 

Now suppose Ax to approach zero, and the point Q to 

approach P. 

Then limit sec QPR = sec TPR = sec <f>. 

limit PQ = ii mit arePQ = limit As * 
Pi2 Pi? Aa; da; 

Hence sec <£ = — ; 
da; 



therefore c ^- = Vl + tan 2 <£ = Jl + /^Y " . (1) 



da; \ ^cta 

It is evident also that 

sin<£ = ^, cos<£ = ^ (2) 

ds ds v y 

It may be noticed that these re- 
lations (1) and (2) are correctly rep- 
resented by a right triangle, whose 
hypothenuse is ds, sides dx and dy, 
and angle at the base <f>. 

Here ds = V (dx) 2 + (dy) 2 , 




TANGENT AND NORMAL. 109 

98^. Differential Coefficient of the Arc. Polar Co-ordinates. 
From the figure of Art. 96, by considering the limiting triangle 
of PMQ, we have. 

limit sec PQB = limit ^§ = limit—, 
RQ Ar 
rU 
or sec^ = — (1) 

dr v J 



Hence 



Vl+tanV^l+r 2 ^) 2 , .... (2) 



^~^i~v + (i) (3) 

It may be noticed that these 
relations (1), (2), and (3), are cor- 
rectly represented by a right tri- 
angle, whose hypothenuse is ds, 
sides dr and rdO, and angle be- 
tween dr and ds, \p. 

Here 

ds = V(dr) 2 +(rd6) 2 , 
and thence 



<Ji+f(¥\\ > 




dr \ \drj d9 



99. Equations of the Tangent and Normal. Having given 
the equation of a curve y =f(x), let it be required to find the 
equation of a straight line tangent to it at a given point. 

Let (x 1 , y') be the given point of contact. Then the equation 
of a straight line through this point is 

y — y'=m(x — x'), (a) 

in which x and y are the variable co-ordinates of any point in 
the straight line ; and m, the tangent of its inclination to the 
axis of X. But since the line is to be tangent to the given 
<curve, we must have, by (1) Art. 9i, 

m = tan cf> — — , 
dx 



110 DIFFERENTIAL CALCULUS. 

-*- being derived from the equation of the given curve y =/(#), 
and applied to the point of contact (x\ y') . 

If we denote this by -2-, we have, substituting m = — in 
dx dx 

equation (a), 

y-y' = %{*-*) (i) 

dx' 
for the equation of the required tangent. 

Since the normal is a line through (#', y') perpendicular to the 
tangent, we have for its equation 

1 fl x i 

y - y, = -^} X - X,) =-^ X - Xl) ' ' ' (2) 
dx' 
For example, find the equations of the tangent and normal to 
the circle x 2 + y 2 = a 2 at the point (x', y') . 

Here, b}~ differentiating x 2 -f y 2 = a 2 , we find 

dy x - , . , dii' x' 

-2- = , from which -—= :• 

dx y dx' y' 

Substituting in (1), we have 

y — y' = — X -(x — x'), 

as the equation of the required tangent. 
It may be simplified as follows : — 
yy' — y' 2 = — xx' -J- x' 2 , 
xx' -\-yy'= x' 2 + y' 2 = a 2 . 

The equation of the normal to the circle is found from (2) 
to be 



x 



which reduces to 



V 1 

x' 



TANGENT ANI> NORMAL. Ill 

EXAMPLES. 

Find the equations of the tangent and normal to each of the 
three following curves at the point (#', y') : 

1 . The parabola y 2 = 4 ax. 

Ans. yy'=2a(x + x'), 2 a (y — y') -\-y'(x — x') = 0. 

2. The ellipse ^ + f* = l. 

a 2 b 2 

Ans. ^ + ^=1, Vx'(y-y') = a 2 y'(x-x'). 

3. The equilateral hyperbola 2xy — a 2 . 

Ans. xy' + yx' = a 2 , y'(y — y') = x\x — x'). 

4. Show that in the preceding curve the area of the triangle 

formed by a tangent and the co-ordinate axes is constant 
and equal to a 2 . 

X s 

5. In the cissoid y 2 = , find the equations of the tan- 

2a — x 

gent and normal at the points whose abscissa is a. 

Ans. At (a, a), y=2x — a, 2y-\-x = 3 a. 
At (a, — a), y-\-2x = a, 2y = x — 8a. 

8 a 3 

6. In the witch y = — -, find the equations of the tangent 

4 a 2 -far 

and normal at the point whose abscissa is 2 a. 

Ans. x-{-2y — 4:a, y = 2x — 3a. 

7. In the curve f— ) -|- f— j = 1, find the equation of the 

w w 

tangent at the point (ar , y') . Ans. °^- + ^ ~*~ w \ — 1 . 

a 2 3b i y ,k 

8. In the ellipse x 2 + 2y 2 — 2xy — x — 0, find the equations of 

the tangent and normal at the points whose abscissa is 1. 

Ans. At (1,0), 2y = x-l, y + 2x=2. 

At (1,1), 2y = x + l, y + 2x = $ 



112 DIFFERENTIAL CALCULUS. 

9. In the parabola x^ + y* = a^, find the equation of the tan* 
gent at the point (x', y') . Ans xx r-\ + yy ,-} _ a l 

10. Show that in the preceding curve the sum of the intercepts 

of the tangent on the co-ordinate axes is constant and 
equal to a. 

2. 2 2. 

11. In the hypocycloid x 3 -\-y 3 = a 3 , find the equation of the 

tangent at the point (x', y'). AnSt xx t~i + yy ^ = t 

12. Show that in the preceding curve the part of the tangent 

intercepted between the co-ordinate axes is constant and 
equal to a. 

100. Asymptotes. Rectangular Co-ordinates. When the 
tangent to a curve approaches a limiting position, as the dis- 
tance of the point of contact from the origin is indefinitely 
increased, this limiting position is called an asymptote. In 
other words, an asymptote is a tangent which passes within a 
finite distance of the origin, although its point of contact is at 
an infinite distance. 

101. From the equation of the tangent (1) Art. 99, we 
find for its intercepts on the co-ordinate axes, 

Intercept on X = x' — y 1 — , 
dy' 

Intercept on F= y' — x 1 -^-- 
dx' 

If either of these intercepts is finite for x' = oo, or y' = oo, 
the corresponding tangent will be an asymptote. 

The equation of this asymptote may be obtained from its 
two intercepts, or from one intercept and the limiting value 



ASYMPTOTES. 113 

102. Omitting the accents in Art. 101 as no longer neces- 
sary, let us investigate the conic sections with reference to 
asymptotes. 

(1) . The parabola, y 2 = 4 ax. 

Here * = ?i 

dx y 

dx 1/ 

Intercept on X=x — y — = x — -*— = —x, 
dy 2 a 

T , dy 2ax y 

Intercept on J — y — x— = y = -• 

dx y 2 

When x = oo, y = <x>, and both intercepts are also infinite. 
Hence the parabola has no asymptote. 



(2). 


The 


hyperbola, - - 


y - 
~b 2 


= 1. 


Here 






dy = 
dx 


_6 2 » 
a 2 y 






Intercept on 


X = 


y 

- i 

X 






Intercept on 


Y = 


: _& 2 

y 



These intercepts are both zero when x = ao, and there is an 
asymptote passing through the origin. To find its equation, 

it is necessary to find the value of — % when x = oo. 

ax 

dy b 2 x L bx ,b 1 






dx a 2 y a ^/ x 2 _ a 2 

Hence — = ± -, when x = oo. 
dx a 



114 DIFFERENTIAL CALCULUS. 

There are then two asymptotes, whose equations are 

y = ±-x. 

a 

(3). The ellipse, having no infinite branches, can have no 
asymptote. 

103. Asymptotes Parallel to the Co-ordinate Axes. When, in 
the equation of the curve, x = oo gives a finite value of y, as 
y = a, then y = a is the equation of an asymptote parallel to X 

And when y = oo gives x = a, then x = a is an asymptote 
parallel to T. 

104. Asymptotes by Expansion. Frequently an asymptote 
may be determined by solving the equation of the curve for x 
or y and expanding the second member. 

For example, to find the asymptotes of the hyperbola 

*?. _ f = l 

a 2 b 2 



y 



a a\ x 2 ] a\ 2ar / 



As x increases indefinitely, the curve approaches the lines 



y= ± — , the asymptotes. 



105. Asymptotes. Polar Co-ordinates. From the figure of 
Art. 97, it is evident that for an asymptote, the polar subtan- 
gent OT has a finite limit, as OP is indefinitely increased. 

clB 
That is, when r 2 — has a finite limit for r=oo, there is an 

dr 
asymptote at that distance from the pole, and parallel to r. 

dB 
If the distance r 2 — is positive, it is to the right, and if 
dr 
negative, to the left, of the pole, looking in the direction of 

the infinite r. 



ASYMPTOTES. 



115 



106. For example, find the 
asymptotes of the curve 

r = a tan 0. 



Here 



dr 
dO 



= a sec 2 0, 



and the subtangent = r 2 — • 



= a sin' 



= ± 



When 
we have r = oo, 
and the subtangent = a. 




There are two asymptotes perpendicular to OX, at the distance 
I 



a from the pole, on each side of it. 



EXAM PLES. 

Investigate the following curves with reference to asymp 
totes : 



J 



x s 



1. y = 

2. y 3 =6x 2 — x 3 . 

3. The cissoid y 2 — 

4. jc 3 + y 3 = a 3 . 

5. (x — 2a)y 2 = x 3 - 



x 3 



2a — x 



Asymptote, y = x. 

Asymptote, x -f- y = 2. 

Asymptote, a? =2 a. 

Asymptote, x-\-y = 0. 

Asymptotes, x = 2 a, a? -f- a = ± 2/. 



6. # 3 + y 3 = 3 acw/. Asymptote, x + ?/ + a = 0. 
(Substitute ?/ = vx in the given equation and in the 

expressions for the intercepts.) 

7. The reciprocal spiral r = — 

Asymptote parallel to OX, at the distance a above. 



116 DIFFERENTIAL CALCULUS. 

8. r = a sec 20. 

There are four asymptotes at the same distance - from the 
pole, and inclined 45° with OX. 

9. The parabola r = - — — — There is no asymptote. 

10. (r~ a) sin = b. 

There is an asymptote parallel to OX, at the distance b 
above. 

11. r = a(sec20 + tan20). 

There are two asymptotes parallel to = -, at the distance 
a on each side of the pole. 



v. 



CHAPTER XIII. 

DIRECTION OF CURVATURE. POINTS OF INFLEXION. 

107. A curve is either concave upward or concave downward. 
It will now be shown that when the equation of the curve is in 
rectangular co-ordiuates, the curve is concave upward or down- 
ward, according as — - is positive or negative. 

ax 

108. Lemma. If u is a function of x which increases as x 

increases, then — - > ; but if u decreases as x increases, — < 0. 
dx dx 

For, in the former case Au and Ax have the same sign, and 
\ 'it du 

therefore — > 0, and consequently — > 0. 
Ax dx 

In the latter case, Au and Ax have different signs, and there- 
fore — < 0, and — < 0. 

Ax dx 




109. By inspection of the first of the two figures above, we 
see that when the curve is concave upward, <f> t7icreases as x 
increases, and consequently tan<£ increases as x increases. 

Hence, by Art. 1 08, d tan ^ > ; 
dx 



lhat is, 



£^ >0 or^>0. 

dx \dxj dx 2 



118 



DIFFERENTIAL CALCULUS. 



From the second figure, we see that when the curve is con- 
cave downward, tan </> decreases as x increases, and therefore 
d tan<£ 



that is, 



dx 2 



dx 
"<0 



<0; 



110. A Point of Inflexion of a curve is a point P, where the 
curvature changes, the curve on one side of this point being 
concave upward, and on 
the other, concave down- 
ward. Hence, by Art. 109, 



dx 1 




at a point of inflexion, 
changes sign ; that is, 

— £ =0 or oo . 

^ X 

It is evident that the tan- 
gent at a point of inflexion intersects the curve at that point. 

Find the point of inflexion of the curve y = (x — l) 3 , and the 
direction of curvature on each side of it. 

Here ^M = 6(x-1). 

dx 2 

Putting this equal to zero, we have for the required point of 



KA/dj 



inflexion, x = 1 . If x < 1 , — * < ; and if x > 

Hence the curve is concave downward on the left, and con- 
cave upward on the right, of the point of inflexion. 



EXAMPLES. 
Find the points of inflexion, and the direction of curvature, 
of the three following curves : — 

1. The curve a' 2 y = — — ax 2 -\- 2 a 3 . 

4a\ 
',, — i ; concave downward on the left of this 

point, concave upward on the right. 



Ans. 



POINTS OF INFLEXION. 119 

2. The witch y= 8a,S ■ 

or -f 4 a 2 

^Lws. f ± , — ] ; concave downward between these 

points, concave upward outside of them. 

3. The curve y~ — 



^ + 3a 2 



Ans. (—3a, ), (0,0), (3a, — J; concave up- 
ward on the left of first point, downward between 
first and second, upward between second and 
third, and downward on the right of third point. 

4. Find the points of inflexion of the curve (-} + [ ^ ) 3 = 1. 



Ans. x = ± — - ■ 

V2 

5. Find the points of inflexion of the curve a A y 2 — a?x* — x e \ 



Ans. x = ±-\27 — 3V33. 



CHAPTER XIV. 



CURVATURE. 



RADIUS OF CURVATURE. 
AND INVOLUTE. 



EVOLUTE 



111. Definition of Curvature. If a point moves in a straight 
line, the direction of its motion is the same at every point of its 
course ; but if its path is a curved line, there is a continual 
change of direction as it moves along the curve. This change 
of direction is called curvature. 

The direction at any point being the same as that of the tan- 
gent at that point, the curvature may be determined by compar- 
ing the linear motion of the point with the simultaneous augular 
motion of the tangent. The curvature is either uniform or 
variable. 

112. Uniform Curvature. The curvature is uniform when, as 
the point moves over equal arcs, the tangent turns through equal 
angles. It is then measured by the angle described by the tan- 
gent while the point describes a unit of arc. 

Suppose the point P to move in the curve AQ. Let s = AP 
denote its distance along the curve from any fixed points, and 
let <£ = PTX, the angle made 
by the tangent PT with the 
fixed line OX. Then as the 
point describes the arc PQ, 
which is denoted by As, the 
tangent turns through the 
ano-le QRK or A<£. Then, if 
the curvature is uniform, it 

is equal to — • 
H As 

The circle is the only curve of uniform curvature. Supposing 

APQ an arc of a circle, if we draw the radii CP and (7Q, and 

let r denote the length of the radius, then the angle PCQ 

^QBK=Ac{>; but arc PQ = CP X angle PCQ; that is, As = rA<f>. 




RADIUS OF CURVATURE. 121 



As 

Hence r = — : that is, the radius of a circle is the reciprocal of 

A<£ 

its curvature. 

113. Variable Curvature. In this case the tangent does not 
turn through equal angles as the point describes equal arcs. 

Here — is the mean curvature throughout the arc As. The 
As 

curvature at the beginning of this arc is more nearly equal to 
— ^, the shorter we take As. Hence the curvature at any point 

is the limit of — ?, that is, — • 
As ds 

114. Radius of Curvature. A circle tangent to a curve at 
any point, and having the same curvature as that of the curve 
at that point, is called the circle of curvature; its radius, the 
radius of curvature ; and its centre, the centre of curvature. 

The curvature of this circle being that of the given curve, is 

equal to — *. If we denote the radius of curvature by p, then 

by Art. 112, P = J^ (1) 

To obtain p in terms of x and y, we have from (1), Art. 98, 



dor, \ \dx 



From (1), Art. 94, tan cf> = % </> = tan" 1 ($M). 

dx \dxj 

tfy_ 
Differentiating, g = -^- 2 (2) 

\dXy 



1 + 



Hence 



ds 
ds dx 
d<f>~ defy dry 

dx dx 2 



dy\ 2n3 
da. 



(3) 



122 DIFFERENTIAL CALCULUS. 



Also, by interchanging x and y, we have 



tfx ' - 

dtf 

which is sometimes the more convenient expression. 

As an example, find the radius of curvature of the semi« 
cubical parabola ay 2 — x 3 . 

tv 4* a.- *• dy 3a£ cl 2 y 3 

Differentiating, -?- = — - 5 - JL = -• 

clx 2a * dx 2 4(aa . )y 

Substituting in (3) , we find 



cc Y (4q + 9a;) 2 
6a 



EXAMPLES. 

Find the radius of curvature of the following curves : — 

1. The parabola y 2 —kax. Ans. p = ^ — i-= — — . 

a? sin 3 <£ 

2. The equilateral hyperbola 2xy = a 2 . Ans. p = ^ "* ^ ' . 

a 2 

3. The ellipse £ + ^= 1. A.. p = W + *£* 

F a 2 6 2 H a *b* 

What are the values of p at the extremities of the major 
and minor axes ? 

4. The curve f-Y+ f~¥= *> at the P oint (°> 6 )- 



5. The curve y = l°g sec x * 

6. The parabola x* + i/ 2 " = a 5 ". -4??s. 



^.ns. 


6 2 , a 2 
— and — 
a b 


,&). 




H 36 


-4?is. 


p = sec x. 


P = 


2(x + y)% 


ai 



RADIUS OF CUBVATUBE. 123 



^ 7. The catenary y = a -(e a -\-e a ). Ans. P = ^ 

8. The hypocycloid X s -J- y* = a*. Ans. p = 3 (ax?/) 3 . 

9. The curve a 4 y 2 = a 2 # 4 — # 6 , at the points (0, 0) and (a, 0). 

^i>is. /o = - and p = a. 

10. The cissoid y 2 = -£—. Ans. p = ax '[ ^ a ~ 3x ) ~. 

* 2a-a; ^ 3(2a-x) 2 

115. Radius of Curvature in Polar Co-ordinates. Resuming 

ds 
(1) Art. 114, p = — , let us express p in terms of r and 0. 



From (3) Art. 981 || = ^Jr 2 + 
From (c) Art. 96, 



* = * + #, ... ^=l+#. 



From (6) Art. 96, 



tan^ = |,or> = tan-|| 



Differentiating, ^ - ^ 



dr\ 2 d 2 r 
dOJ 

7* + 



dxj, _\dOJ r d0 2 



d0, 

Substituting, ~ B 

w 



124 



DIFFEBENTIAL CALCULUS. 



EXAM PLES. 
• Find the radius of curvature of the following curves : — 



1 . The circle r = a sin 6. 

2. The logarithmic spiral r = e° a . 

3. The spiral of Archimedes r = ad. 

4. The cardioid r = a (1 — cos 6). 

A 

5. The curve ?*=asin 3 -- 

3 

6. The parabola r = a see 2 — 

7. The lemniscate r 2 = a 2 cos 2 6. 



Ans. o = 



Ans. 



Ans. 



rVl+a 2 
(r 2 + ct 2 ) f 



r 2 + 2a 2 
Ans. p 2 = -ar, 



Ans. p = -a sin 



,0 



Ans. p = 2a sec 3 - 
H 2 

Ans. p = — 
H 3r 



116. Co-ordinates of the Centre of Curvature. Let x, 2/ be the <Q 
co-ordinates of P, any 
point of the curve AB, 

and C the corresponding «*\ n j 

centre of curvature. CP 
is then the radius of 
curvature, and is normal 
to the curve. 

Draw also the tangent 
PT. 

Then CP=p\ 

angle PCR = PTX = <f>. 

Let a, /?, be the co-ordinates of C. 

OL = 0M- RP, LC = MP + BC; 
that is, a = x — psin4>, fi = y + pcoscj> . . . (1) 

To express a and ft in terms of x and y, we have, by (2) Art. 
98, and (1), (2), Art. 114, 




EVOLUTE AND INVOLUTE. 



125 



. , ds dy dy dy dx 
d<j> ds dcf> dx dcj> 



ds dx dx \dx 

p COS d> = = — — = 7T- 

^ dcfids dcj> d 2 y 



dy 
dx 



dx, 



Hence 



dy 
dx 



1 + 



1 



dry 
dx 2 



dx 2 



£ = 2/ + 



dry 
dx 2 






H 



d?y 
dx 2 






(2) 



117. Evolute and Involute. Every point of a curve AB has a 

g corresponding centre of 
curvature. Thus, P 1? P 2 , 
P 3 , etc., have for their 
/ ^-respective centres of 
ir* curvature d, 2 , (7 3 , etc. 
p 2 The curve .HiT, which 
is the locus of the cen- 
tres of curvature, is 
X" called the evolute of AB. 
To express the inverse relation, AB is called the involute of HK. 




118. To find the equation of the evolute of a given curve. 

By (2) Art. 116, a and /?, the co-ordinates of any point of the 
required evolute, may be expressed in terms of x and y, the 
co-ordinates of any point of the given curve. These two equa- 
tions, together with that .of the given curve, furnish three 
equations between a, /?, x, and y, from which, if x and y are 
eliminated, we obtain a relation between a and (3, which is the 
equation of the required evolute. 

For example, find the equation of the evolute of the parabola 
y 2 = 4 ax. 

dy l -i dry 1 1 _f 

dx dx~ 2 



Here 



126 



DIFFERENTIAL CALCULUS. 



Substituting in (2) Art. 116, we have 
a= Sx-{- 2a 

Eliminating x, we have for 
the equation of the evolute, 

aB 2 = ~ (a -2 a) 3 . 
r 27 v J 

This curve is the semi- 
cubical parabola. The figure 
shows its form and position. 
F is the focus of the given q 
parabola. 

00= 2a =2 x OF. 

119. Properties of the Invo- 
lute and Evolute. Let us return 
to the equations, (1) Art. 116, 

a = x — p sin<£, 

/3 = y + P cos 0. 
Differentiating with reference to s, and by (2) Art. 98, and 
(1) Art. 114, we have 

da 

ds 

dfi _dy dp 




ds ds ds ds 



. . , dd> dp , 

cos d> — p sin <p — - = — ^-cos <£ . 
ds ds ds ds ds 

Dividing (6) by (a), 



(a) 
(6) 



P 



— £- =s — cot <& = tan [ d> + 

da ^ V 2 , 

If <£' denote the angle made with the axis of X by the tan- 
gent to the evolute, then, by (1) Art. 94, 



-*- = tan<£\ 
da 



.-. <£'=<£ + 



That is, the tangent to the evolute is perpendicular to the 
corresponding tangent to the involute. In other words, a tan- 
gent to the evolute at any point Ci (Fig. Art. 117), is C x Pn the 
normal to the involute at P x . 



EVOLUTE AND INVOLUTE. 127 

120. Again, from (a) and (6), Art. 119, 



^dsy \dsy \dsj \ds J \ds / 

where s' denotes the length of the arc of the evolute measured 
from a fixed point. Hence, 

— ■ = ± -£ , and therefore As'= ± Ap. 
ds ds 

That is, the difference between any two radii of curvature 
PiCi, P3O3, is equal to the corresponding included arc of the 
evolute C1C3. 

121. From the two properties of Arts. 119 and 120, it fol- 
lows that the involute AB may be described by the end of a 
string unwound from the evolute HK. From this property the 
word evolute is derived. 

It will be noticed that a curve has only one evolute, but an 
infinite number of involutes, as may be seen by varying the 
length of the string which is unwound. Such curves are called 
parallel curves. 

EXAMPLES. 

1. Find the co-ordinates of the centre of curvature of the 

cubical parabola y s = a?x. 

6a 2 y H 2a* 

2. Find the co-ordinates of the centre of curvature of the 

X X 

catenary y = - (e a -+- e a ). 

Ans. a = x — - V2/ 2 — a 2 , (3 = 2y. 

3. Find the co-ordinates of the centre of curvature, and the 

o 2 

equation of the evolute, of the ellipse — + %- = 1. 

a 4 b A 

(aa)^ + (&/3)$ = (a 2 -& a )i 



128 DIFFEBENTIAL CALCULUS. 

4. Show that in the parabola x^ + y* = a? we have the rela- 

tion a -f (3=3(x + y). 

5. Find the co-ordinates of the centre of curvature, and the 

equation of the evolute, of the hypocycloid X s + y^ = a 9 . 

Ans. a = x-\-3x 3 y f , (3 = y + 3x*y 3 , 
(a + /3)* + (a-/3) f = 2a f . 

6. Given the equation of the equilateral hyperbola 2 xy = a 2 ; 

show that 

a- a 2 

Thence derive the equation of the evolute 

(a + /3) f -(a-/?) f = 2a f . 



CHAPTER XV. 



ORDER OF CONTACT. OSCULATING CIRCLE. 




Fig. 1. 



122. Definition. Suppose two curves to have two common 
points P 1 and P 2 . If one 
of these points, as P 2 , be sup- 
posed to approach to coinci- 
dence with P 1} the limiting 
position is called a contact 
of the first order. Thus two 
curves are said to have contact 
of the first order when they 
have two consecutive common 
Doints. 

Again, suppose the two 
curves, having at P a con- 
tact of the first order, to have 
a third common point P 3 . Now 
when P 3 moves up to coinci- 
dence with P, we have ulti- 
mately a contact of the second 
order, which thus denotes three 
consecutive common points. 

Similarly, suppose the two 
curves to have three common 
consecutive points at P, form- 



ing a contact of the second 




order, and a fourth common 
point P 4 . By supposing P 4 
to move up to P, we have a contact of the third order, contain- 
ing four consecutive common points. 

In general, a contact of the ?ith order includes n -f- 1 consecu- 
tive common points. 



130 DIFFERENTIAL CALCULUS. 

123. When the order of contact is even , the curves, cross at 
the point of contact; but ivhen the order is odd, they do not 
cross. " 

For a contact of the first order, it is evident from Fig. 1, 
Art. 122, that outside of P x and P 2 , the dotted curve is on the 
same side of the other curve. Hence, when the two points 
coincide to form the point of contact, the curves do not cross 
at that point. 

For a contact of the second order, it is evident from Fig. 2, 
Art. 122, that when P 3 coincides with P, the curves cross at 
the point of contact. 

For a contact of the third order, Fig. 3, Art. 122 shows that 
the curves do not cross at the point of contact. 

Similarly it is evident that the proposition is generally true. 

124. Osculating Curves. As a straight line can be made to 
pass through only two points, the tangent has generally a 
contact of only the first order with a curve. 

The circle having the closest contact with a curve at a given 
point is called the osculating circle. As a circle can be made 
to pass through only three points, the osculating circle has 
generally contact of the second order. 

The parabola of closest contact is likewise called the oscu- 
lating parabola. As a parabola can be made to pass through 
four points, the osculating parabola has contact of the third 
order. 

The conic of closest contact is called the osculating conic. 

As a conic can be made to pass through five points, the 
osculating conic has contact of the fourth order. 

It is evident from Art. 123 that the osculating circle and 
osculating conic cross the curve at the point of contact, while 
the tangent and osculating parabola do not. 

125. Exceptional Points. Although the tangent has gener- 
ally contact of the first order, it may at exceptional points of 
a curve have a contact of a higher order. 



OBBEB OF CONTACT. 



131 



For example, since the tangent at a point of inflexion crosses 
the curve, it follows from Art. 123, that the order of contact 
must be even. Hence at a point of inflexion the tangent has 
contact of at least the second order. 

The osculating circle, which has generally contact of the 
second order, has a higher order of contact at points of maxi- 
mum or minimum curvature, as, for example, the vertices of 
an ellipse. It is evident from the symmetry of the ellipse 
with reference to its vertices, that no circle tangent at these 
points would cross the curve at the point of contact. Hence, 
by Art. 123, the order of contact is odd, — at least the third. 



126. Analytical Conditions for Contact. 

Let y = cj>(x), and y = i]/(x), 

be the equations of two curves having two common points 
P and Q. 

Y 




O a-" M , X 

Let OM= a, MN= h. 

Then <£(a) = i/r(a), and <f>(a + li) = ij/(a + h). 

Expanding each member of this equation by Taylor's 
Theorem, 

<£ (a) + h<f>' (a) + 1 </>" (a) + 1 <£"' (a) + • • • 






(1) 



132 



DIFFERENTIAL CALCULUS. 



Since cj>(a)=if/(a), we have from (1) after dividing by h, 

+ '(a).+ |+»(a) + ...=*'(o) + |^(o) + .... 

When Q approaches P, h approaches zero, and we have at 
the limit , , , N . , f . 

Hence the conditions for a contact of the first order at the 
point x = a, are 



127. Again, suppose the two curves have a contact of the 
first order at P and another common point Q. 






As before, let Oilf = a, MN— h. 
Since <£(a) = i//(a), and <£'(«) = ^'(a), 

we have from (1) Art. 126, after dividing by A 2 , 

When Q approaches P, we have at the limit, when h = 0, 

*"(a) = ^(a). 

Hence the conditions for a contact of the second order at the 
point x — a, are 

*(a)=<Ka), *'(a) = <fr'(«)> *» = ^"(a). 



OSCULATING CIRCLE. 133 

128. Conditions for contact of the nth order. The same 
process may be extended to contacts of higher orders, every 
additional point in the contact adding one to the series of 
equalities at the end of the preceding article. 

In general, the conditions for a contact of the nth order at 
the point x = a, are 

<!>(a)=+(a), *'(a)=f (a), *"(a)=^"(a), »• *»(a)=^(a). 

In other words, for x= a, 

dy fy_ dry 

y ' dx dx 2 ' dx n ' 

must all have the same values, respectively, taken from the 
equations of both curves. 

129. To find the co-ordinates of the centre, and radius, of the 
osculating circle at any point of a given curve. 

Let the equation of the given curve be 

y =/(«)• 

The general equation of a circle with centre (a,b) and 
radius r, is 

(x-a) 2 +(y-b) 2 =r 2 (1) 

Differentiating twice successively, we have 

x-a + (y-b)^=0, (2) 

1+ (2)V-»>S- a ^ (3) 

From (3), y-b = ^L. ....... (4) 



From (2), x-a = 



dx 2 

\dx 
dx 2 



1+/* 



)'} 



(5) 



134 



DIFFERENTIAL CALCULUS, 



Substituting (4) and (5) in (1). 



Jz 



\dxj 



Hence 



a — x 



dx 2 

dy 
dx 



HiJ] 



dry 
dx 2 



and 



Ml 



dx 2 



6 = 2/4- 



mi 

dry 
dx 2 



In these expressions, 



dx* dx 2 " 1 



(6) 



(7) 



(8) 



refer to (1), the equa- 



tion of the circle ; but since the osculating circle by definition 
has contact of the second order with the given curve, these 
quantities will have the same values if derived from the 
equation y=f(x), at the point of contact. 

By comparing (7) and (8) with the expressions for a, j3, 
and p, in Arts. 114, 116, it is evident that the osculating 
circle is the same as the circle of curvature. 



130. At a point of maximum or minimum curvature, the 
osculating circle has contact of the third order. 

Tf we regard equation (8) in the preceding article as re- 
ferring to the given curve, y =f(x), we have as a condition 
for a maximum or minimum value of r, 

a> 
dx 

We thus obtain from (8), 



0. 



(See Art. 146.) 



3<fy(<tfy\ 

dx [dx 2 J 



2 



1 + 



df 2 ' 

dx, 



3k -0, 

dx> 



OSCULATING CIRCLE. 135 

, , . , d 3 y dx\dx 2 J m 

from which J=__ . 

Again, if we regard (8) as referring to the osculating circle 
(x-af+iy-bf^r 2 , 

we shall also have — = 0, 

dx 

since r is constant for all points on the circle. 

Thus we obtain, both for the curve and the circle, the same 

expression (1) for J, and since -^ and — ^ in the second 
dx 3 dx dx 2 

member of (1) have, at the point of contact, the same values 
for both curves, it follows that — has likewise the same 



dx 3 



value. Hence the contact is of the third order. 

EXAMPLES. 

1. Find the order of contact of the two curves, 

y = x 3 , and y = Sx 2 — 3# -f- 1. 

By combining the two equations, the point, x = l, y = 1, 
is found to be common to both curves. 

Differentiating the two given equations, 
y = x 3 , 

c ?y = 3x 2 , 
dx 

^ = 6x, 
dx 2 

^ = 6, 

dx 3 dx 3 



y = 


= 3a 2 - 


- 3x 


+ 1, 


dx 


: 6X - 


•3, 




d?y_ 
dx 2 


= 6, 






d 3 y _ 


= 0. 







136 DIFFERENTIAL CALCULUS. 

When x — 1, — - = 3, in both curves ; 
dx 

d"u 
when x = l, -~ = 6, in both curves ; 
dxr 

but — ^ has different values in the two curves. 
dx 3 

Hence the contact is of the second order. 

2. Find the order of contact of the parabola Ay = x 2 , and 

the straight line y = x — l. Ans. First order. 

3. Find the order of contact of 

9y = a?-3x 2 +27, and 9?/ + 3« = 28. 

Ans. Second order. 

4. Find the order of contact of 

y = log(x — 1), and x 2 —6x-+-2y + 8 = 0, 
at the common point (2, 0). Ans. Second order. 

5. Find the order of contact of the parabola &y = x 2 — 4 ? and 

the circle x 2 + y 2 - 2 y = 3. Ans. Third order. 

6. What must be the value of a, in order that the parabola 

y = x-\-l-\-a(x — l) 2 , 
may have contact of the second order with the hyperbola 
xy = 3x — l? Ans. a = — 1. 

7. Find the order of contact of the parabola 

(x-2a) 2 +(y-2ay=2xy, 
and the hyperbola xy = a 2 . Ans. Third order. 



CHAPTER XVI. 

ENVELOPES. 

131. Series of Curves. When, in the equation of a curve, 
different values are assigned to one of its constants, the result- 
ing equations represent a series of curves, differing in position, 
but all of the same kind or family. 

For example, if we give different values to a in the equation 
of the parabola y 2 = 4 ax, we obtain a series of parabolas, all 
having a common vertex and axis, but different focal dis- 
tances. 

Again, take the equation of the circle (x— a) 2 -\-{y— b) 2 =c 2 . 
By giving different values to a, we have a series of equal 
circles whose centres are on the line y == b. 

The quantity a which remains constant for any one curve 
of the series, but varies as we pass from one curve to another, 
is called the parameter of the series. 

Sometimes two parameters are supposed to vary simultane- 
ously, so as to satisfy a given relation between them. 

Thus, in the equation of the circle (x — a) 2 -f- (y — b) 2 = c 2 , 
we may suppose a and b to vary, subject to the condition, 

a 2 + b 2 = k 2 . 

We then have a series of equal circles, whose centres are on 
another circle described about the origin with radius Jc. 

132. Definition of Envelope. The intersection of any two 
curves of a series will approach a certain limit, as the two 
curves approach coincidence. Now, if we suppose the param- 
eter to vary by infinitesimal increments, the locus of the ulti- 
mate intersections of consecutive curves is called the envelope 
of the series. 



138 



DIFFERENTIAL CALCULUS. 



133. The envelope of a series of curves is tangent to every 
curve of the series. 

P Q 




Suppose L, M, N to be any three curves of the series. P is 
the intersection of M with the preceding curve L, and Q its 
intersection with the following curve N. 

As the curves approach coincidence, P and Q will ultimately 
be two consecutive points of the envelope, and of the curve M. 
Hence the envelope touches M. 

Similarly, it may be shown that the envelope touches any 
other curve of the series. 



134. To find the equation of the envelope of a given series of 

curves. 

Before considering the gen- 
eral problem let us take the 
following special example. 

Required the envelope of 
the series of straight lines 
represented by 

. m 
y = ax + —, 

a 

a being the variable param- 
eter. 

Let the equations of any 
two of these lines be 




and 



O0J + -, 

a 



y = (a + h)x + 



m 



a-\-h 



a) 

(2) 



ENVELOPES. 139 

From (1) and (2) as simultaneous equations, we can find 
the intersection of the two lines. Subtracting (1) from (2), 

hm 



= hx 



a(a-\-h)' 



or = a; — (3) 

a(a + h) v ' 

From (3) and (1), we have 

x= m , y=( 2a + h )™, ... (4) 

which are the co-ordinates of the intersection. 

Now if we suppose h to approach zero in (4), we have for 
the ultimate intersection of consecutive lines 

m 2m 

a* a 

By eliminating a between these equations we have 

y 2 = i mx, 

which, being independent of a, is the equation of the locus of 
the intersection of any two consecutive lines ; that is, the equa- 
tion of the required envelope. 

The figure shows the straight lines, and the envelope which 
is a parabola. 

135. We will now give the general solution. 

Let the given equation be 

f(x, y, a) = 0, 

which, by varying the parameter a, rep/esents the series of 
curves. 

To find the intersection of any two curves of the series, we 
combine 

f(x,y, a) = 0, (1) 

an <* /(», ft a + 7i) = (2) 



140 DIFFERENTIAL CALCULUS. 

From (1) and (2), we have 

f(x, y,a + h) -f(x, y,a) _ () /qx 

and it is evident that the intersection may be found by com- 
bining (1) and (3), instead of (1) and (2). 

When the two curves approach coincidence, h approaches 
zero, and we have, by Art. 10, for the limit of equation (3), 

~-f(x 9 y 9 a) = (4) 

Thus equations (1) and (4) determine the intersection of 
two consecutive curves. By eliminating a between (1) and 
(4) we shall obtain the equation of the locus of these ultimate 
intersections, which is the equation of the envelope. 

136. Applying this method to the preceding example, 

. m 
y = ax -f — , 

a 
we differentiate with reference to a, and obtain for (4) Art. 135, 

= x - -• 

a 2 

Eliminating a between these equations gives the equation 
of the envelope, 

y 2 = 4mcc, as before. 



137. The evolute of a given curve is the envelope of its 
normals. 

This is indicated by the figure of Art. 117, and the proposi- 
tion may be proved by the method of Art. 135, as follows : 

The general equation of the normal at the point (x', y') is 
by (2) Art. 99, 

x-x< + ^(y-y>) = 0, (1) 



ENVELOPES. 141 

in which the variable parameter is as', the quantities y', -^-, 

dx' 

being functions of x'. Differentiating (1) with reference to 
x'j we have 

- 1 -(SJ + ^-^S=° < 2 > 

From (1) and (2) we find for the intersection of consecu- 
tive normals, 



dx' 2 



x~x' — 



dx' 



dx 



T\ 



dx' 2 



As these expressions are identical with the co-ordinates of 
the centre of curvature in Art. 116, it follows that the envelope 
of the normals coincides with the evolute. 



EXAMPLES. 

1. Find the envelope of the series of straight lines repre- 

sented by y = 2m# + m 4 , m being the variable param- 
eter. 
Differentiating the given equation with reference to m, 

= 2a-f4m 3 . 

Eliminating m between the two equations, we have for 
the envelope, 

16y 3 + 27x* = 0. 

2. Find the envelope of the series of parabolas y 2 =a(x— a), 

a being the variable parameter. ^ ?ls# 4w 2 = a£ 



142 DIFFERENTIAL CALCULUS. 

3. Find the envelope of a series of circles whose centres are 

on the axis of X, and radii proportional to (m times) 
their distance from the origin. ^[ ?lS- y*— m 2 (x 2 -±-y 2 ). 

4. Find the evolute of the parabola y 2 = 4 ax according to 

Art. 137, taking the equation of the normal in the form 

y = m (x — 2 a) — am 3 . Ans. 27 ay 2 — 4 (x — 2 a) 3 . 



x 2 v 2 

5. Find the evolute of the ellipse — + ^ = 1, taking the 

equation of the normal in the form 

by = ax tan <£ — (a 2 — & 2 ) sin $, 
where <£ is the eccentric angle. 

Ans. (axf -f (byf= (a 2 - b 2 f. 

6. Find the envelope of the straight lines represented by 

sccos30 -\-ysm30 = a(cos20)2, 
6 being the variable parameter. 

Ans. (x 2 + y 2 ) 2 = a 2 (x 2 — y 2 ), the lemniscate. 

7. Find the envelope of the series of ellipses, whose axes 

coincide and whose area is constant. 
The equation of the ellipses is 

± + t-\ (1) 

a 2 + 6 2-1 ' K) 

a and b being variable parameters, subject to the con- 
dition ab = k 2 , (2) 

calling the constant area irk 2 . 
Substituting in (1) the value of b from (2), 

s+f =*> < 3 > 

in which a is the only variable parameter. Differen- 
tiating (3) with reference to a, we have 



ENVELOPES. 143 

__2tf 2af =0 

a 3 ^ k 4 K } 

Eliminating a between (3) and (4), we have 

4afy 8 = ft*. 

Second Solution. Differentiate (1), regarding both a and 
b as variable. 

of da , y 2 db A /c .. 

i^ + V = (5) 

Differentiating (2) also, we have 

bda + adb = Q (6) 

From (5) and (6), we have 

t = t (7) 

a 2 b 2 K } 

From (7) and (1), 

a 2 b 2 2 v; 

Substituting (8) in (2), 

4 cc 2 ?/ 2 = k\ 

8. Find the envelope of the circles whose diameters are the 

double ordinates of the parabola y 2 = 4 a«. 

ulns. ?/ 2 = 4 a (a + jc) . 

9. Find the envelope of the straight lines - -f V. = 1, 

when a n -\-b n — k n . jl _?l jl_ 

-4ns. & n+1 + y 1l+1 = k n+ \ 

o»2 rt.2 

10. Find the envelope of the ellipses — h — , = 1, 

a- b- 

when a + b = k. Ans. a* + y* = kK 



144 DIFFERENTIAL CALCULUS. 

11. Find the envelope of the circles passing through the 

origin, whose centres are on the parabola y 2 = 4 ax. 

Ans. (x + 2 a) ?/ 2 + x 3 = 0. 

12. Find the envelope of circles described on the central radii 

of an ellipse as diameters, the equation of the ellipse 

being ^ + ^=1. Ans. (x 2 + y 2 ) 2 = aV + b 2 y 2 . 

a o 

13. Find the envelope of the ellipses whose axes coincide, and 

such that the distance between the extremities of the 
major and minor axes is constant and equal to k. 

Ans. A square whose sides are (x ± yf = k 2 . 



CHAPTER XVII. 

SINGULAR POINTS OF CURVES. 

138. The term singular points is applied to points of a curve 
having some peculiar property independent of the position 
of the co-ordinate axes. 

We proceed to consider the different varieties of singular 
points. 

Points of Inflexion. These have already been considered 
in Art. 110. 

Multiple Points. These are points through which several 
tranches of a curve pass. The figures show a double point 
and a triple point. 





139. To find the multiple points of a curve. It is evident 
that at such a point there are several tangents, and therefore 

-£ has more than one value. 
dx 

Suppose the equation of the curve, free from radicals, to be 

f(x,y) = 0. 
Then by (2) Art. 67, we have 
du 
dy dx 

di=~dZ' where u =K*>y)- 

dy 



146 DIFFERENTIAL CALCULUS. 

Since u contains no radicals, this expression for -^ can 

dx 

have bnt one value at any given point, unless it takes the 
form - ; that is, 

^=0, and ^=0 (1) 

dx dy 

These are therefore the conditions for a multiple point. 
If values of x and y which satisfy (1) also satisfy the 
equation of the curve f( \ _ n 

J\ x ) y) — V} 

we have for any such point 

dy = 
dx 

This indeterminate form can be evaluated by the method 
of Art. 53. 

The result of the process of evaluation will be an equation 

of the second, or higher, degree with respect to -^, thus 

dx 

determining several values of that quantity. This will be 
apparent from an example. 



140. Let us examine for multiple points the lemniscate 

(x 2 -\-y 2 ) 2 =a 2 (x 2 -y 2 ). 
Here u = (x 2 + y 2 ) 2 + a\y 2 - X s ) = 0. 

— = ±x(x i + y 2 )-2a 2 x, 

dx 

d ^ = ±y(x> + y*)+2o?y. 
dy 

Putting ^ = 0, and — = 0, 

dx dy 

we find x = 0, y = 0, or x = ± — -, y = 0. 



SINGULAR POINTS. 



147 



Of these values of x and y, x—Q, y = 0, alone satisfy the 
equation of the given curve. Let us find the value of 

-^ for this point. 
dx r 



du 
dy _dx _ 2x 3 +2xy 2 -a 2 x _ 
dx ~ du ~~ 2x 2 y + 2y z +d 2 y ~~ q 

dy 



, when x = 0, y = 0. 



Evaluating by Art. 53, 



Ay 



6x 3 +22/ 2 +4x?/^-a 2 
dy dx — or , A „ 

-^ = — = — — , when x = 0, y = 0. 

dx 4^ + (2* 2 +6/+a 2 )^ a ^ 

dx dx 



Hence 



\dx) 



or 



ax 



The origin is a double point, the two tangents being inclined 
45° to X. 

Y 




141. Again, take the curve whose equation is 
u = x 4 -\- 2 a# 2 i/ — ay s = 0. 



3w , o . , du 

— = 4 x 6 -f- 4 aajy, — 
d# a?/ 



= 2 ax 2 -3 a?/ 2 . 



148 



DIFFERENTIAL CALCULUS. 



(ill f)l/ 

Putting — = 0, and — = 0, we find a; = 0, y = 0, to be the 
ox dy 

only point of the curve satisfying these conditions. 

In finding the values of -2, 
ax 

let tfi=-~ 9 and y*=-7?- 

ax ax' 

3 ay 2 — 2 ax 2 
Evaluating by Art. 53, 

\2x 2 +kay -\-A,axv x -, a a 

y 1= _T ^JE *- 1 =-, whena = 0, y = 0. 

bayy 1 —4:ax 



Evaluating again, 

= 24,x + 8ay l -\-Aaxy 2 



Say-, 



6«2/i -f- 6a2/2/ 2 — 4a 0>ay 2 — 4a' 



, when x = 0, 2/ = 0. 




Hence Vii^Vi- 2) =4y D 

and therefore 2h=0, or y 1 =±^/2. 

Hence the origin is a triple point as shown in the figure. 



SINGULAR POINTS. 



149 



142. Points of Osculation. A multiple point is called a 
point of osculation when the branches of the curve passing 
through it are tangent to each other. 

In this case — will have two or more equal value at the 
• . dx 

point. 

For example, consider the curve 



a*y 2 = a 2 x 4 — x 6 . 



Here 



du 
dx 



= -4aV + 6aj B , 



u = aV— a¥+ x Q = 0. 
du 



dy 



= 2 ay 



4a 2 ar 5 -6ar 5 , a a 



Evaluating by Art. 53, 



yi= — tttt. — =7T-z-> wnen x =°> y=& 



2 ay 



2 ay 




Hence 2ay 2 =0, giving two values of y 1 = 0. The origin 
is a point of osculation. 



150 



DIFFERENTIAL CALCULUS. 



143. Cusps. When the branches of the curve are only on 
one side of the point of osculation, this point is called a cusp, 
as Pi or P 2 . 





tf= 


X s . 


y = 


3 

±x* 9 


dy _ 


±U. 



The conditions for a cusp are the same as those for a point 
of osculation, with the additional condition of imaginary 
points of the curve on one side of this point. 

For example, take the semicubical 
parabola 



Here 



dx 

When x = 0, ^ = ± 0. 
dx 

There are then two coincident 
tangents at the origin. But since 
y is imaginary for negative values 
of x, there are no points on the left 
of the origin. Hence the origin is 
a cusp. 




144. Conjugate Points. If, in determining a multiple point, 
the values of _? are imaginary, we then have a point of the 

QjX 

curve through which no branches pass ; that is, an isolated 
point. Such a point is called a conjugate point. 



SINGULAB POINTS 
For example, the curve 

ay 2 — x 3 -f- bx 2 = 0, 



151 





dy 
dx 


3^-2 
2 ay 


bx 
~0' 


Hence 








dy 
dx 


6oc 

2 


-2b 

a c M 
dx 


b 
dx 


len x-. 


= o, 


y = 0. 




Therefore 






dy 
dx 


= ± 







gives 
when x = 0, y = 0. 



Hence the origin is a con- 
jugate point. This appears 
directly from the given equa- 
tion 

ay 2 — x 2 (x — b), 

from which it is evident that 
besides the origin, there are 
no points of the curve when 
x<b. 

EXAMPLES. 

1. Show that the curve 

A. 

a-y l = a*x z — or, 
has a multiple point at the origin. 

2. Show that the curve 

y 2 = xlog(l + x), 
has a multiple point at the origin. 




152 DIFFERENTIAL CALCULUS, 

3. Show that the cissoid 

o Xi 

y =o > 

Z« — X 

has a cusp at the origin. 

4. Show that the curve 

x 3 + 2x? + 2xy - y 2 + 5% - 2y = 0, 
has a cusp at the point ( — 1,-2). 

5. Show that the curve 

(a* + ff = a¥ + by, 
has a conjugate point at the origin. 

6. Show that the curve 

ay 2 = (x — af(x — b), at the point (a, 0)^ 
has a conjugate point, if a < b ; 
a double point, if a > b ; 

and a cusp, if a = b. 



CHAPTER XVIII. 



T 50D 



MAXIMA AND MINIMA OF FUNCTIONS OF ONE 
INDEPENDENT VARIABLE. 

145. Definition. A maximum value of a function is a value 
greater than those immediately preceding or immediately following. 
A minimum value of a function is a value less than those 
immediately preceding or immediately following. 

If the function is represented by the curve y = f(x), then 

PM represents a 
p i maximum value of 

y or of /(#), and 
QJSF represents a 
minimum value. 

146. To find 
the conditions for 
a maximum or a 
minimum. 
X It is evident that 
at both P and Q the tangent is parallel to the axis of X, and 
therefore we have as a condition for both maxima and minima, 




dy 

dx 







(a) 



Again, at P the curve is concave downward, and at Q, con- 
cave upward. 

Hence, by Art. 109, 



d 2 v 
for a maximum value, — - < 0, 

dx 2 

for a minimum value, — ^> 0. 

dx 2 



Mmtvn f \ 



ilL 



(6) 



For example, find the maximum and minimum value of 



Put 



2/ = ^_2a; 2 + 3£+l. 



154 



DIFFERENTIAL CALCULUS. 



Then 



dx dx 2 



2 x -4 



By (a), a 2 -4^ + 3 = 0. 
Solving this equation, 
cc = 1 or 3. 
To apply (b), we substitute both x = 1 and a; = 3 in 

2cc-4, 



do; 2 



and find 



when x = 1, — < 0, 
dec 2 

when cc = 3, — ^ > 0. 
dx 2 



Hence when x = 1, ?/ is a maximum ; 

when # = 3, 2/ is a minimum. 
The maximum value of 2/ is 2\, and the minimum value, 1. 



147. In exceptional cases it may happen that the value of 

/-72 

x given by (a) makes — ^ = 0, so that neither of the con- 

(XX 

ditions (b\ is satisfied. This 
would be the case for a point 
of inflexion R, whose tangent 
is parallel to OX. Here the 
ordinate EL is neither a 
maximum nor a minimum. 

But there may be a maxi- 
mum or minimum value of 

d 2 v 
v -y, even when — ^ = 0. This 
* ff} dx 2 

is more fully considered in Art. 150. The method of the 
following article is also applicable to such cases. 




148. Second Method of determining Maxima and Minima. 
Referring to the figure of Art. 145, and supposing x to 



MAXIMA AND MINIMA FOB ONE VARIABLE. 155 

increase, we see that as we approach. P, y increases, and on 
leaving P, y decreases. Hence, by Art. 108, -^ is positive on 

ax 

the left, and negative on the right, of P. That is, when y 

is a maximum, -^ changes from -f- to — . 
dx 

Similarly, it may be shown that when, as at Q, y is a 

minimum, -2 changes from — to -f . 
dx 
These relations may also be obtained by noticing that tan <£, 

which is equal to -*- 9 changes sign at P and Q. 
ax 

Let us apply these conditions to the example in Art. 146, 
where 

^ = »■- 4s + S= («-!)(» -3). 

dx 

Here — can change sign only when x = 1 or x = 3. 

dx 5 5 J 

By supposing x to be first slightly less, and then slightly 
greater, than 1, we find that x — 1 changes from — to + ; 

but since x — 3 is then negative, it follows that — changes from 

dx 

+ to — , when x = 1, and denotes a maximum. In the same 

way, we find that -^ changes from — to -f- , when x — 3, and 
dx 

denotes a minimum. 

Again, consider the function y=(x — 4) 5 (# + 2) 4 . 

Here & = 3 (3 x - 2) (x - 4) 4 (x + 2) 3 . 
ax 

When x = -, -^ changes from — to -f ; 
3 dx 

when x = — 2, -£ changes from + to — ; 
' dx 5 ' 

when aj = 4, -^ does no£ change sign, 
dx 

since (cc — 4) 4 cannot be negative. 



156 



DIFFERENTIAL CALCULUS. 



Hence we conclude that y is a minimum when x = - ; a max- 

o 

imum when x = — 2 ; but neither a maximum nor minimum 

when x = 4. 

As this method does not require 3— 9 , it is preferable to that 
of Art. 146, when the second differentiation of y involves much 
work. 

149. Case ivhere -^=00. It is to be noticed that — some- 
dec dx 

times changes sign by passing through infinity instead of zero. 



Hence if 



dy _ 
dx 



for a finite value of x, this value should be examined, as well 
as those given by -^ = 0. 

(XX 

For example, suppose 

y = a — b{x — cy. 



Then 



dM = - 



2b 



dx 3(x-c)*' 




hence we have 

-2 = 00, when aj = c. 
dx 

It is evident that when 



-^ changes from 



cfo; 



+ 



to — , indicating a maximum 
X value of y, which is a. 

The figure shows the max- 
imum ordinate PM, corre- 
sponding to a cusp at P. 



MAXIMA AND MINIMA FOR ONE VARIABLE. 157 



On the other hand, suppose y = a — b(x — of 

dy = ^_ 

dx 



Then 



S(x-cf 



— = oo, when x = c. 



cfy 



But as -^ does not change sign when a? = c, there is no 
da? 
maximum nor minimum. The corresponding curve is shown 
in the figure. 

Y 




150. Conditions for Maxima and Minima by Taylor's 
Theorem. Suppose the function f(x) to be a maximum when 
x = a. Then, by the definition in Art. 145, 



and also 



f(a)>f(a + h) } 
f(a)>f(a-h), 



where h is any small but finite quantity. Now, by the sub- 
stitution of a for x in Taylor's Theorem, we have 

/(a + ft)-/(a)= ft/'(a) + ^/''(a) + ^/"'(a) + ... (1) 
f(a - h) -/(a) = -^'(a) + f/» _-|/'''(a) + ... (2) 

By the hypothesis /(a -f 7i) —f(a) < 0, 
and also /(a — &) —f(a) < 0. 






158 DIFFERENTIAL CALCULUS. 

Hence the second members of both (1) and (2) must be 
negative. 

By taking h sufficiently small, the first term can be made 
numerically greater than the sum of all the others, involving 
h 2 , h 3 , etc. Thus the sign of the entire second member will 
be that of the first term. As these have different signs in (1) 
and (2), the second members cannot both be negative unless 

/'(«) = 0. 
Equations (1) and (2) then become 

f(a + h) -/(a) = j|/"(a) + 1/"'(«) + - 
f(a - h) -/(a) = |/"(a) - |/'»(a) + .... 

The term containing h 2 now determines the sign of the 
second members. That these may be negative, we must ha^e 

/"(tt)<0. 

If then /' (a) = and /" (a) < 0, 

f(a) is a maximum. 

Similarly, it may be shown that if 

/'(a) = and /"(o)>0, 

f(a) will be a minimum. 

If f'(a)=0 and /"(a) = 0, 

similar reasoning will show that for a maximum we must also 

have /»'(a) = and / iv (a)<0; 
and for a minimum 

/"'(o) = and / iv (a)>0. 

151. The conditions may be generalized as follows : 
Suppose 

/'(a)=0, /"(a) = 0, /'"(a) = 0, -. /»(a) = 0. 



MAXIMA AND MINIMA FOE ONE VARIABLE. 159 

. Then if n is even, /(«) is neither a maximum nor a 

minimum. 

If n is odd, f(a) will be a maximum or minimum, according 

as 

/ n+1 (a)<0 or >0. 



EXAMPLES. 



-7 



1. Find the maximum value of ax — x 2 . Ans. — , when x = -• 

4 J 2 

2. Find the maximum and minimum values of 

2x 3 — 9x 2 + 12x — 3. Ans. x = l gives a maximum, 2 ; 

# = 2 gives a minimum, 1. 
^ 3. Find the maximum and minimum values of 

x 3 —3x 2 —9x-\-5. Ans. x = — 1 gives a maximum, 10; 
x = 3 gives a minimum, — 22. 

/4. Show that ar 3 — Sx 2 + 6# has neither a maximum nor min- 
imum value. Cvmm» - - 

5. Show that ax -4- - ? is a minimum, when cm? == - = -y/ab. 

a/ a; 

a 2 b 2 

6. Show that the least value of ■ 1 is (a + b) 2 . 

sin 2 6 cos 2 

Investigate the following functions for maxima or minima : 

,- _ x 2 — lx-\- 6 Ans. x = 4t gives a maximum value of y ; 
x — 10 x = 16 gives a minimum value of y. 

Ans. A minimum when x = e. 



8. 


y = 


X 






logic 




9- 


y = 


(s- 


a)(b- 


-X) 



, 2ab . • -, (a-&) 2 

Ans. x = erives a maximum value, -^ - 1 — 

a + b 5 4«6 

10. y = 2tanx — tan 2 #. ^Ins. A maximum when »; = -• 

4 



160 DIFFERENTIAL CALCULUS. 

11. y = sinse(l -f-cosce). Ans. A maximum when x = ^ 

o 

12. y = tana; -f 3 cot x. Ans. A minimum when x = ^ 

o 

13. 2/ = sin a; cos (x — a). Ans. A maximum when aj = - + -i 

a minimum when a; = • 

2 4 

14. 2/ = ^ — ~ ' • Ans. A minimum when x = -» 

a — 2a; 4 

15. 2/ = (a;-l) 4 (« + 2) 3 . 

5 

^tis. A maximum when x = ; a minimum when se = 1 ; 

7 
neither when a; = — 2. 

16. y = (x-2) 5 {2x + iy. 

Ans. A maximum when x = ; a minimum when a; = — ; 

neither when x — 2. 

17. 2/ = (^ + l) f (^-5) 2 . 

Ans. A minimum when a; = 5 ; a maximum when x = - ; 
a minimum when cc = — 1. 

18. # = (2a;-a)3(a;-a)^ 

2a 
Ans. A maximum when a; = — ; a minimum when a; = a. 

3 ' 

PEOBLEMS IN MAXIMA AND MINIMA. 

1. Divide 10 into two such parts that the product of the 
square of one and the cube of the other may be the greatest 
possible. 

Let x and 10 — x be the parts. Then a; 2 (10— a;) 3 is to be a 
maximum. Letting u = a^(10 — a;) 3 , we find 

— = 5z(4 - a;) (10 - a;) 2 = 0, 
dx 

from which we find that it is a maximum when x = 4. Hence 
the required parts are 4 and 6. 



MAXIMA AND MINIMA FOE ONE VARIABLE. 161 

2. A square piece of pasteboard whose side is a, has a small 
square cut out at each coruer ; find the side of this square that 
the remainder may form a box of maximum contents. 

Let x = the side of the small square. Then the contents of 
the box will be (a — 2x) 2 x. Eepresenting this by u, we find 
that u is a maximum when x = -, which is the required answer. 



3. Find the greatest right cylinder that can be inscribed in 
a given right cone. 

Suppose the figure to be a section 
through the axis AD. 

Let AD = a, DC=b. 

Let x — DQ, the radius of the base 
of the cylinder, and y=PQ, its alti- 
tude. 

From the similar triangles ADC, 
PQC, we find 



— ^ — ■ = - or y = -(o 
b-x b * b x 



x). 




The volume of the cylinder is 



7rarfy = 



7r-x 2 (b — x). 
b 



This will be a maximum when u=bx 2 — x 3 is a maximum. 
This is found to be when x = -§-&, the radius of the base of the 
required cylinder. 

From this, y = -, the altitude of the cylinder. 
o 

4. Determine the right cylinder of the greatest convex 
surface that can be inscribed in a given sphere. 

Suppose the figure (page 162) to be a section through the 
axis of the cylinder, AB. 

Let r 5= OP, the radius of the sphere. 

Let x = OB, the radius of the base of the cylinder, 
and y =PB, one-half its altitude. 



162 DIFFERENTIAL CALCULUS. 

From the right triangle OPM we have 

** • • («) 

The convex surface of the cylinder 
is 



x 2 + y 2 




2irX'2y = 47rcc Vr 2 — a^ 



= 47rVrV— a; 4 . 

This will be a maximum when 
u = t^x 2 — a? 4 is a maximum. 

This is found to be when x = -^-, 

V2 
the radius of the base of the required 
cylinder. 

From this, 

rV2. 

Another solution of the problem is the following : 
Since the convex surface is 4tirxy f put u = xy, to be a maximum. 
du 



Hence the altitude of the cylinder is 



y+x c }y z= o. 

dx dx 



But from (a), x + y 



dy 

dx 



0. 



(<0 



Eliminating -^ from (&) and (c), we have x = y, which, 
dx 

combined with (a), gives the same result as before. 

5. A rectangular piece of pasteboard 30 inches long and 14 
inches wide has a square cut out at each corner; find the side 
of this square so that the remainder may form a box of maxi- 
mum contents. Ans. 3 inches. 

6. Divide a into two parts such that the product of the mth 
power of one and the nth power of the other may be a maxi- 
mum. Ans. The required parts are proportional to m and n. 

7. A person being in a boat 3 miles from the nearest point 
of the beach, wishes to reach in the shortest time a place 5 miles 



MAXIMA AND MINIMA FOR ONE VARIABLE. 163 

from that point along the shore ; supposing he can walk 5 miles 
an hour, but row only at the rate of 4 miles an hour, required 
the place he must land. 

Ans. One mile from the place to be reached. 

8. The top of a pedestal which sustains a statue 11 feet high 

is 25 feet above the level of a man's eye ; find his horizontal ~ - 
distance from the base of the pedestal when he sees the statue 
subtending the greatest angle. Ans. 30 feet. 

9. Through a point (a, 6), referred to rectangular axes, a 
straight line is to be drawn, forming with the axes a triangle 
of the least area. Show that its intercepts on the axes are 2 a 
and 26. * 

10. Through the point (a, b) a line is drawn such that the 
part intercepted between the axes is a minimum. Show that its 
length is (a* + &*)*. 

^^11. Given the slant height a of a right cone ; find its altitude 
when the volume is a maximum. . a 

ns ' vt' 

12. Given a point on the axis of the parabola y 2 = 4:<xx 1 at 
the distance h from the vertex ; find the abscissa of the poiut 
of the curve nearest to it. Ans. x = h — 2a. 

13. Find the maximum rectangle that can be inscribed in an 
ellipse whose semi-axes are a and b. 

Ans. The sides are aV2 and ?>V2 ; the area, 2ab. 

14. A rectangular box, open at the top, with a square base, 
is to be constructed to contain 108 cubic inches. What must 
be its dimensions to require the least material ? 

Ans. Altitude, 3 inches ; side of base, 6 inches, 

15. Find the altitude of the right cylinder of greatest volume 

inscribed in a sphere whose radius is r. A 2r 

* Ans. — =;. 

V3 



164 DIFFERENTIAL CALCULUS. 

16. Find the altitude of the right cylinder inscribed in a 
sphere whose radius is r, when its entire surface is a maximum. 

Am. (2-— Vr. 






Vfe. 

17. Find the altitude of the right cone of greatest volume 
inscribed in a sphere whose radius is r. Ans. ±r. 

18. Find the altitude of the right cone of maximum entire 
surface inscribed in a sphere whose radius is r. 

Ans. (23 - VT7) ^j* 

19. Find the altitude of the right cone of least volume cir- 
cumscribed about a sphere whose radius is r. 

Ans. Its altitude is 4r, and its volume is twice that of the 
sphere. 

20. Find the altitude of the least isosceles triangle cir- 
cumscribed about an ellipse whose semi-axes are a and 6, the 
base of the triangle being parallel to the major axis. 

Ans. 3 b. 

21. A tangent is drawn to the ellipse whose semi- axes are a 
and 5, such that the part intercepted by the axes is a minimum. 
Show that its length is a + b. 

22. The lower corner of a leaf, whose width is a, is folded 
over so as just to reach the inner edge of the page. Find the 
width of the part folded over, when the length of the crease is 
a minimum. Ans. fa. 

23. In the preceding example, find when the area of the tri- 
angle folded over is a minimum. 

Ans. When the width folded is § a. 







CHAPTER XIX. 

MAXIMA AND MINIMA OF FUNCTIONS OF TWO OR 
MORE INDEPENDENT VARIABLES. 

152. Definition. A function, f(x, y), of two independent 
variables has a maximum value, when 

/(»> y) >/(« + *> y +■*)> 

for all small values of h and 7c, positive or negative j and a 
minimum value, when 

/fcJO </(* + *>? + *)• 

153. Conditions for Maxima or Minima. 
Letting u =/(», y), 

we have from Art. 68, 

f(x + h,y + 7c) -f(x, y) = h-^ + k^ 

In order that u may be a maximum, the second member of 
(1) must be negative for small values of h and 7c, positive or 
negative. By similar reasoning to that in Art. 150, it is 
evident that the sign of (1) is determined by the terms con- 
taining the lowest powers of h and 7c ; that is, by 

, du , 7 du 

7i— + 7c~— 

ox ay 

Hence, in order that (1) may not change sign with 7i and 7c, 
we must have 

j die , t du p. 
7i— + 7c-— = 0. 
dx dy 



166 DIFFERENTIAL CALCULUS. 

As 7i and k are independent of each other, this is equivalent to 

£-* ^ g=° < 2 > 

Equation (1) then becomes 
fix + h,y + k) -fix, y) = \(Ah 2 + 2Bhk+Ck*) + ..., 

where A = b \ B = ^L, C = ^- 

dx 2 dxdy dy z 

But Ah 2 + 2 Bhk + Ck 2 = ( Ah +^) 2 + (^C-B?)k\ m (3) 

A 

In order that (3) may preserve the same sign for all small 
values of h and k, it is necessary that AC — B 2 should be 
positive ; for if negative, the numerator of (3) will be positive 
when k=0, and negative when AJi -j- Bk = 0, Hence we have 
as an additional condition for a maximum, 

B 2 <AO (4) 

The sign of (3) then depends upon that of the denominator 
A. Hence for a maximum we must have 

A<0. 

Similarly it may be shown that for a minimum value of u 7 
we must have (2) and (4), together with 

A>0. 

It may be noticed that (4) requires that A and C should 
have the same sign. Hence if A is positive, C will be also. 
The exceptional cases, where 

B 2 =AC, 

or where A=0, B = 0, (7=0, 

require further investigation. We shall not consider them 
here. 



MAXIMA AND MINIMA FOR TWO VARIABLES. 167 

154. The conditions for a maximum or minimum value of 
u = f(x, y), may be restated as follows : 
For either a maximum or minimum, 



also 



du A -, du r, 

— =0, and — = 0; 
ox dy, 



/d 2 ttV dhi dhz 
\dydxj dx 2 *■ 



dif 



For a maximum, — - < 0, and — - < 0. 

dy 2 



For a minimum, 



dhi 
dtf 



(1) 
(2) 
(3) 



dx 2 



155. Functions of Three Variables. A similar investigation 
to that in Art. 153, gives as the conditions of a maximum or 
minimum vulue of u =f(x, y, z): — 
' For either a maximum or minimum, 

^ = o, *» = 0, ^ = 0, 

dx dy dz 



Ld 


/ d 2 u\ 2 d 2 udhi 
\dxdyj dx 2 dy 2 


For a maximum 


f^<0, and A<0; 

dx? 


r a minimum, 


^>0, and A>0; 

dx 2 






dhi dhi dhi 






dx 2 dxdy dxdz 


where 


A = 


dhi dhi dhi 
dxdy dy 2 dydz 

dhi dhi d 2 u 
dxdz dydz dz 2 



168 DIFFERENTIAL CALCULUS. 

EXAMPLES. 

1. Find the maximum value of 

w = 3 axy — cc 3 — 2Z 3 - 

Here — = 3 ay — 3 x 2 , — = 3 ax — 3 y 2 . 
dx By 

Also P^-6x, £j = -6y, *L = S* 

dar By 2 BxBy 

Applying (1) Art. 154, we have 

ay — x 2 — 0, and ax — y 2 = ; 
whence a? = 0, y = ; or x = a } y = a. 
The values a; = 0, 2/ = 0, give 

which do not satisfy (2) Art. 154. 
Hence they do not give a maximum or minimum. 
The values x = a, y = a, give 

d 2 w a d 2 u a d 2 u o 

— - = — o a, — - = — ba, • = 3 a, 

dx 2 By 2 BxBy 

which satisfy both (2) and (3), Art. 154. 
Hence they give a maximum value of u which is a 3 . 

2. Find the maximum value of xyz, subject to the condition 

-,+g+S-i a) 

a 2 b 2 c 2 
Fromd), J-1-3-& 

and as xyz is numerically a maximum when afyV is a 
maximum, we put 



MAXIMA AND MINIMA FOB TWO VARIABLES. 169 

dx u \ a? Vf dy \ a? V )' 

52 »_ 0,^/1 6a 2 y"\ d'u _ „ , A x 1 6y 2 \ 

a?" H x "* *> s7 (, * T?f 

3x3?/ ^V « b J 

From — = and — = 0, we find, as the only values 
dx By 

satisfying (2) Art. 154, 

a b -,. -, . 

x = — -, y = , which give 

V3 V3 

a^ = _85_ 2 £^ = _8a_ 2 &u__ _4a6 

dx 2 9 ' dy 2 9 ' 6a% 9 

As these values satisfy (2) and (3), Art. 154, it follows 
that xyz is a maximum when 





a b 


, 2 = 


c 

V5 




The maximum value of xyz is ' 


abc 
3V3 




3. 


Find the values of x, y, z, that render 






x 2 -\- y 2 + z 2 -\- x - 


■2z — xy 






a minimum. A 




2 



-| 3 2/ = ~q' 2==1 - 



4. Find the maximum value of 



(a — a;)(a — y)(a? + y — a). -4ws. — • 



5. Find the minimum value of 

x 2 -\- xy -f- y 2 —ax —by. 



Ans. ^{ab — or—b 2 ), 
o 



170 DIFFERENTIAL CALCULUS. 

6. Find the values of x and y that render 
sin a; -f- sin?/ + cos (x + y) 



a maximum or minimum. 



q 

Ans. A minimum, when x = y = — ; 

a maximum, when x = y = -- 

6 

7. Find the maximum value of 

(aa?-f- fry -4- c) 2 . 2 . ,« . * 

^+r+i 

8. Find the maximum value of a^V, subject to the condition 

2x + 3y + 4z = a. An ^ fa 



?)' 



9. Divide a into three parts such that their continued 
product may be the greatest possible. 

Let the parts be x, y, and a — x—y. 

Then u = xy (a — x — y), to be a maximum. 

f4 ■) i fill 

— = ay — 2 xy — y 2 = 0, — = ax — x 2 — 2 xy = 0. 
dx ay 

These equations give x = y = -- 

o 

Hence a is divided into equal parts. 

Note. — When, from the nature of the problem, it is 
evident that there is a maximum or minimum, it is 
often unnecessary to consider the second differential 
coefficients. 

10. Divide a into three parts, x, y, z, such that x m y n z p may 
be a maximum. x v z a 

' m n p m+n +p 



MAXIMA AND MINIMA FOB TWO VARIABLES. 171 

11. Divide 30 into four parts such, that the continued prod* 

uct of the first, the square of the second, the cube 
the third, and the fourth power of the fourth, may 
be a maximum. Ans. 3, 6, 9, 12. 

12. Given the volume a s of a rectangular parallelopiped; find 

when the surface is a minimum. 

Ans. When the parallelopiped is a cube. 

13. An open vessel is to be constructed in the form of a 

rectangular parallelopiped, capable of containing 108 
cubic inches of water. What must be its dimensions to 
require the least material in construction ? 

Ans. Length and width, 6 in. ; height, 3 in. 

14. Find the co-ordinates of a point, the sum of the squares 

of whose distances from three given points, 

Oi, Vi), (%2, 2/ 2 ), (a> 3 , Vs), 
is a minimum. A 1 , x 1 , . N 

the centre of gravity of the triangle joining the given 
points. 

15. Find the volume of the greatest rectangular parallelopiped 

that can be inscribed in the ellipsoid 

<£+£ + * = !. Ans. 1252- 

« 2 6 2 c 2 SV3 



INTEGRAL CALCULUS. 

CHAPTER I. 

ELEMENTARY FORMS OF INTEGRATION. 

1. Definition of Integration. The inverse operation of dif- 
ferentiation is called integration. By differentiation we find 
the differential of a given function, and by integration we find 
the function corresponding to a given differential. This func- 
tion is called the integral of the differential. 

For instance, 

since 2xdx is the differential of x 2 , 

therefore x 2 is the integral of 2xdx. 

The symbol j is used to denote the integral of the expres- 
sion following it. 

Thus the foregoing relations would be written, 



d(x 2 ) = 2xdx, I 2xdx=x 2 . 



It is evidently the same thing whether we consider this 
integral, as the function whose differential is 2xdx, or the 
function whose differential coefficient is 2x. 

As a matter of notation, however, it is not customary to 
write 

I 2 x = x 2 , but always | 2 xdx = x 2 . 

Integration is not like differentiation a direct operation, but 
consists in recognizing the given expression as the differential 



174 INTEGRAL CALCULUS. 

of a known function, or in reducing it to a form where such 
recognition is possible. All functions can be differentiated, 
but all cannot be integrated 5 that is, their integrals cannot 
be expressed in terms of known functions. 

2. Elementary Principles. 

(a). It is evident that we have 

I 2xdx — x 2 + 2, or j 2xdx — x 2 — 5, 

as well as J 2xdx = x 2 \ 

since x 2 + 2 and x 2 — 5 are functions, each of whose differen- 
tials is 2xdx. 



In general j 2xdx — x 2 + c, 



where c denotes an arbitrary constant called the constant of 
integration. 

Every integral in its most general form includes this term, 
+ c. (We shall omit this constant of integration in the folio w- r _. u 
Ing integrals, as it can readily be added when necessary, r^ QrVw 

(&). Since d{u ± v ± iv) = du ± dv ± dw y 

it follows that 

I {du ± dv ± dw) = \du± j dv ± I dw. 

That is, we integrate a polynomial by integrating the sepa- 
rate terms, and retaining the signs. 

(c). Since d(au) =adu, 

it follows that I a du = a I dw. 

That is, a constant factor may be transferred from one side 
of the symbol j to the other, without affecting the integral. 



ELEMENTARY INTEGRALS. 175 

3. Fundamental Integrals. We shall now give a list of 
formulae, which may be regarded as fundamental, and to 
which all integrals must ultimately be reduced. We shall 
then consider in this chapter such examples as are integrable 
by these formula^ either directly, or after some simple trans- 
formation. 

I. Cu n du = ^—~. 
J n + 1 

II. j — = log u. 
J u 

III. Ca u du = -2— . 

J log a 

IV. Ce u du = e u . 

V. J cos udu = sin u. 

VI. f sinudu = — cosu. 

VII. I sec 2 udu = tanw. 

VIII. I cosec 2 ttdw = — cotw. 

IX. j sec u tan udu = sec w. 

X. j cosecwcotw<2w = — cosecw. 

XI. j tan. u du — log sec u. 

XII. J cot udu = log sin w. 

XIII. j sec udu = log (sec u -f- tan it) = log tan ( - -f 7 ) • 

XIV. j cosec udu = log(cosecw — coti<) = logtan^. 



176 INTEGRAL CALCULUS. 

XV. I— — — = -tan -, or = 

J u 2 -+- or a a a 

vTrr C du 1 -, w — a 1 , a — w 

XVI. I- 5 = —log— ■— , or =__log— — - 

J w — a- la u4-a la a-\-u 



or =_-cot- 1 -. 



XVII. f du = sin- 1 g. or = - cos" 
J Va 2 - u 2 a 

XVIII. f dtt = log (u + V« 2 ± a 2 ) . 

J Vir ± a 2 

vtv C du 1 1« 1 

XIX. I =-sec~ 1 - ? or = 

J u^ u 2 — a 2 ci a a 



cosec -1 



Vw 2 -a 2 a a a a 

du l u 

- = vers -• 

■\/2au-u 2 a 



w C du iM 

XX. I — = vers 2 - 

•J ■ 



4. Proof of I. and II. 

To derive L, 
since d(w M+ J) = (n -f- l)w n cZu, 

therefore 

w »+i = C( n + 1) tt n dM = ( n _j_ -^ jv^ ? ^ ( c ) Art> 2. 

w n dw = • 

n + 1 

Formula II. follows directly from 

71 du 

alogw = — 
u 

It is to be noticed that I. applies to all values of n except 
n = — l. For this value, it gives 



J u -1 du = 
Formula II. provides for this failing case of I. 



u v 

- = CO. 



ELEMENTARY INTEGRALS. 177 

EXAMPLES 
For Formulae I. and II. 

Integrate the following expressions : 

/X 5 
x 4 dx = — , by L, where u = x, and n = 4. 

2. C(x* + l)$xdx. 

If we apply L, calling u — x 2 + 1, and n = - ; then 

We must then introduce a factor 2 before the #$#, and 
consequently its reciprocal - on the left of | . 

C(x 2 + l)*a>efo = 1 C(x 2 + 1)* 2 ajefo, by (c) Art. 2. 

Hg' + l)* ' (a* + l)* 

2 3 3 * 

2 

J ^-3^ 3 J a; 3 -3a 2 a3 

= 1 log (x? - 3 a 2 a) = log (ic 3 - 3 a 2 x)K 
o 

By introducing the factor S, we make the numerator the 

differential of the denominator, and then apply II. 



4 



f (V + Sx 8 - 6x ll )dx = -(x 6 + 2a; 9 - 3a; 12 ). 

- Cx 2 - 1 7 ar 2 , 

7. I efcc = log a;. 

J x Z 

8. f-£*L = iogVra. 

Jr-1 



178 INTEGRAL CALCULUS. 

J xr-\-2x 
in C(x 2 -2fdx 2 6 , a? , 6 

. I (a 2 — ar) 3 V£cfa = 2#2 — — . 1. 

J V ; \3 7 ^ 11 15J 

12. f(V5- V5)^ = tti»- 2 aa* + ^-^ 

13. f(a? + l) 2 da? = fo + 1 ) 3 . 

Integrate also, after expanding (a; + 1) 2 . How are the two 
results reconciled ? 

14 r(^ n - « w y^g = .g.^ __ 4 a n) + t^i ogak 

%J X AlTi 



35. ^-2^ + 2)^-1)^= ^ -2^ + 2) t 



Integrate also, after multiplying x 2 — 2x-\-2 by a — 1, and 
compare the two results. 

16. f (3 ax 2 - a 3 )* (2 aa; - ar°) da? = — (3 ax 2 - a^)i 

Integrate also, after multiplying numerator and denominator 
by 2, and compare the two results. 

18 - /-%,= (»*)"• 

(nx) « 

'19. f ^"" 1 ~ 1 da; = - log (af - na?) . 

J x n — nx n 

20. f(--|= - vf=W = § E(« s + *¥ ~ (« 2 + *fo 

*' Wcr + ar va" + ar/ ° 



ELEMENTARY INTEGRALS. 179 



2L ff^i dx = x - los(2x+3) °- 

23. f ** =4(Va+Vi)i 

J a;* 

25. Jl^±l)^ = l [ l 0g(a; + 1)]2 . 



+ 
(a; + a)* + (a; + &)* 3(a-6) 



26. f ^ - = - 2 - x [(« + o)l-(a? + ft)f]. 



27. f(aj 3 + l)(^ + 5)^a; = -(a) 3 + 5)t. 

Suggestion, (x 3 + 1) (cc 3 + 5)* = (af^ -f 5 ajt )i (arr- + x~?). 

28 r(^+i)<te = g (aP+B jv 

{X n + il)'* 

1 

Suggestion. Multiply numerator and denominator by x n ~\ 

The following integrals may be evaluated by I., after mul« 
tiplying the binomial under the radical sign by x~ 2 . 

= -~ C(a 2 x~ 2 - l)~i ( - 2 a?x- 3 dx) 
z (l %j 

1 (aSar 8 -!)* Q 2 ar 2 - 1)* 
2a 2 1 " a 2 

2 

V« 2 — a; 2 



180 INTEGRAL CALCULUS. 



30 C ^ x = — V ^ + a 2 t 

J ^Var^ + a 2 °? x 



33. 



/ 



Va 2 — x*dx _ (a 2 — as 2 )^ 

a? " 3a¥ 

V& 2 — a 2 dx _ (x 2 — a 2 )f 
ai* ~ 3a¥ 
da; _ a; 

(a 2 — a: 2 )^ a 2 Va 2 — ar* 



J ( 



(z 2 + « 2 )f aWx 2 + a 2 



35. 



/; 



dx V 2 ax — x 2 



x V2 ax — a^ aaJ 



36 ( * flfifa? _ a; 

^ (2ax — x 2 )i aV2ax—x 2 

37. (^ V2 aa; — x 2 dx _ (2 ax — cc 2 )! ^ 

^ x 3 ~ 3aa^ 

qq / aa/ a? — a 

J {2 ax — x 2 )% a? ^/2o,x—x 2 

This may be obtained from Ex. 33 by substituting x—a 
for x. 

5. Proof of III. awe? IV. These are evidently obtained 
directly from the corresponding formulae of differentiation. 

EXAM PLES 
For Formula III. and IV. 



i. fr+w-l^+Jg). 

*• I (e m + e a )dx = — + ae a . 
•s a 

( a - _ 6«) dx = -£ f— r 

nloga ralog& 



ELEMENTARY INTEGRALS. 181 

4. f( e * + e ~ x fdx = he 2 * - e~ 2x ) + 2x. 

5. ri£!±lI 2 ^ = A r ^ + 2e-lY 
•J Ve x Ve x V 3 / 

6. f(3^-l)h*te = l(3e*'-l)*. 

7 ' /S = f + ^ x + 1 °g(^- 1 )- 

8. f^zldx = log(e* + l) 2 - x. 
»/ e x + 1 

9. I a x e x dx = - — 

J 1-f-loga 



10 . f(* x - bx y d x= axb - x - a - xbx -2x. 

J a x b x 



o 

log a — log 6 



6. Proof of Y.— XIV. It is evident that V.— X. are obtained 
directly from the corresponding formulae of differentiation. 
To derive XI. and XII., . 

tan udu = — I — ■ == — log cos u = log sec u. 

J cosu 

cot u du = I = log sin u. 

J sintt 

To derive XIII. and XIV., 

/, f sec u(taa.u + sec u)du rsecutanudu 
secudu = I * '- — = I 
J secw + tanw J secu-± 

= log (sec u + tan u) . 

J, rcosectt(— cotw -f-cosectO^< 
cosec udu = I * ! L — 
J ccsecu — cotw 

= log(cosecw — cotw). 



du +sec 2 u du 
tan-i* 



.182 INTEGRAL CALCULUS. 

By Trigonometry, 

., 2 sin 2 7; 

1 — cos u 2 . u 

cosec u — cot u = — : = = tan-. 

smw . u u 2 

2 sin - cos - 
2 2 

If we substitute in this - -f u f or u, 
2 ; 

we have sect* +tanw = tanf - + -Y 

\4 2/ 

Hence we obtain the second forms of XIII. and XIV. 

EXAMPLES 
For Formulae V.— XIV. 

1. I (sin 2 x -f- cos 2 a) dec = - (sin 2 # — cos 2 a). 

J™ £C 1 

(cos sin 3 a) dec = 3 sin - + - cos 3 x. 
o o o 

3. f[sm(a+bx) -cos( a -ta)]cfo= si "(«~^) -cosfa+to) , 

4. pm3x^ = l sec3g/ 
*/ cos 2 3a; 3 

5. J sec-(sec| + tan^)d^ = 2(tan^-}-sec^ ). 

/» ri — cosao?, 1, , , 

6. I ax = - (cosec ax — cot ax). 

J sin 2 ax a 

7. I (tana; -J- cot x) 2 dx — tan x — cot a?. 

8. I (secic — tan#) 2 cfoc = 2(tana; — sec a;) — a?. 

9. I = log (a -\-b cos x). 

J a + 6cosa; b 

-q r tan a: da; _ log(acos 2 a; + 5sin 2 a;) T 
J a -f- b tan 2 x 2{b — a) 



ELEMENTARY INTEGRALS. 183 

1. I (tan2oj — l) 2 dx = -ta.n2x + logcos2x. 

2. J (sec 2 a; + 1) 2 da; = - tan 2 a; -{-log (sec 2 a; -f- tan2a;) -\-x. 

3. f (coseca: — l)(cotoj+l)c?a; = — ic— coseca?— log(l + coso;). 

4. I (secx4-coseca?) 2 d£c = tana; — cotic + 21ogtano;. 

5. jsin 2 icc?« = sin2#. 

J 2 4 

6. ( cos 2 a;c?a; = -4" -sin2x. 
J 2 4 

7. f 1 + s * ng? cfa = 2(seca? + tana;) - x. 
J 1 — sin a; 

o /•cotsc + tana? , 1, ., fie , \ 

8. I — da; = -logtan[ - + x ). 

J cot a; — tan a; 2 \4 J 

9. ft a nsta n (s + a)ifa = -«- lo g( 1 -- timatanfl, >. 

c/ tan a 

20. ( sec a; sec (a; -fa) da; = 'log • 

J sma cos(a;4-<x) 



7. Proof of XV. -XX. 

To derive XV., 

dw , Az 

d 



/dw __1 /* a -^ C \ a ' _^t -i tt 



To derive XVIL, 

dit 
dw 



/aw r a . .u 

. — | . == sin- 1 -. 

V a 2 — w 2 »/ L u 2, a 



,11 
vers -. 



184 INTEGRAL CALCULUS. 

To derive XIX., 

du 

/du , 1 r a 1 ,u 
= - I =-sen~ 1 -. 
M V*r - a 2 aJ u \v?__ 1 « « 

To derive XX., 

du 

/du r a 

^/2au — u 2 J Lu u 2 
Va~tf 

Since tan -1 - = ^ — cot -1 - , 

a 2 a 

it is evident that a" tan -1 - = d( — cot -1 - 

a \ a 

Hence either expression may be used as the integral in XV. 

In the same way we obtain the second forms of XVII. and 
XIX. 

The formulae XVI. and XVIII. are inserted in the list of 
integrals, because they are of similar form to XV. and XVIL, 
respectively, with different signs. 

To derive XVI., 



1 = 1/ 1 _ 1 \ . 

u 2 — a 2 2 a\u — a u + aj' 
hence 

du 



J w — a- 2 a J \u — a 



u -f-a 



1 1 

= — -[log(u — a) — log(w + a )] = 7T~-^ 



2a 2a u-\-a 

Or we may integrate thus : 



/ du 1 Cf —du du \ 

u 2 — a 2 2 a J \a — u a-\-u) 



— [log (a - u) - log(a + ™)]= — log^— -^. 
2 a 2 a a + u 



ELEMENTARY INTEGRALS. 

To derive XVIII., 
assume s/u 2 ± a 2 = z, a new variable. 

Then u 2 ± a 2 = z 2 , 

2udu=2zdz; 
du dz du -f- dz 



185 



therefore 
Hence 
that is, 



u-\-z 



/du fdu + dz t , , x 



I 



du 



-\/u 2 ± w 



= log (u + vV ± a 2 ) . 



EXAMPLES 
For Formulae XV. -XX. 



-j; 



do; 



-tan -1 - — • 
9aj 2 + 4 6 2 

f cto = 1 3^-2 , 
J9x 2 -4: 12 S 3a; + 2 

3. f jm 

J a/1 _ 



sin -1 2cc. 



V1-4CC 2 



f da; = ^log(2a;4-Vl4-4fl; 2 ). 
^Vl+4x 2 2 



a? da; 



Ji'-4 8 6 ar + 2 
9. f_ 



VI - ^ ^ 

a? da? 1 , _. x 2 

= -tan — 

x* + 4 4 ,2 



da; 1 _n2a; 

= -sec 1 — -. 

#V4a; 2 — 9 3 3 



da? 



a/6^ 



- = vers -• 
a 3 



186 INTEGRAL CALCULUS. 



dx 1 ,2b 2 x 
10. I — - = -vers I 

■Vax - b 2 x 2 & « 



da; 1 _i ace 

=z-sec L — 

Va¥ - b 2 b & 

da? _ 1 

V2a;-3a; 2 V3 



12. f- — - = — vers- 1 3 a;. 

i3. f da? = :-iog^- a = J-iog to + a . 

•/ a 2 — frV 2 2 a& bx-\-a 2ab bx — a 

14. r 2 ^~ 5 ^ = -Iog(3a^ + 2)--^-taii- 1 — ♦ 
J3a; 2 + 2 3 8V T ; V 6 V6 

1K /~2a; — 5,. l n /Q - ON 5 , ar\/3— V2 

15. I dx = -log(3x 2 — 2) log — — 

J 3x^-2 3 8k ; 2V6 *VS+V2 



The same formulse may be applied to expressions involving 
x 2 + aa; + o or — x 2 + ax + b, by completing the square 
with the terms containing x. Thus, — 



dx 1, iSB + l 

= -tan — ! — ■ 

x + iy + 4 2 2 



16. f ** = f- 

Jse* + 2a; + 5 J ( 



17 r da? = r 2da? = r 

J V2 + x - x 2 * V8 + 4a;-4ar 2 J 



2dx 



sin" 



V9-(2z-l) 5 
i2a;-l 



18 -f^ 



dx 1 , _,x — 3 
— = tan *■ 

6a; + 11 V2 V2 



/da; _ 1 -J x — 5 
a^_6a; + 5~4 ° g o> - 1* 

r (to 1 lQg 2a^ + 3-V5 

J 3^ + 3^+1 V5 2a; + 3+V5 

21. f *? =ltaa-*5*=i 

J5a^-2a; + l 2 2 



20 



ELEMENT ABY INTEGRALS. 187 



— ** = s in- 1 ^^. 






3x-x 2 Vl3 

23. f- da? = log(a;-2+vV-4a; + 13), 

= sec a tan -1 (# sec a — tana). 

a 2 -2asilla-j-l V 

25. f da; =— log(3a?-2 4-V9g 2 -12a?). 
J V'3x 2 -4;x V3 



26. f ^ =llog 3 ^ 

J 3ar*-f lOa + 3 4 S 

cZa; = 2_ 

oa^ -I- bx + c V4ac-6 2 "~ V4ac-& 2 ' 



3^-f IOoj + 3 4 a + 3 

_ ^ = - ? tan- 1 2a * + 5 



••/; 



1 , 2aa;-f-&— V& 2 -4ac 

or =• — log — 

V& 2 — 4ac 2ax-\- b -|-V& 2 — 4ac 
_1_ 

Vax 2 -+- bx -f- c Va 



28. r__ -^— __ = JL .log (2 ax + b + 2 VaVaar 2 + &e + c> 

•/ -\/ n.crr' -X- hof. -J- /? -v//7. 



c?cc 1 . i 2ax—b 
29. I = — -sm • 



'•fe 



V— aic 2 -}- 6x + c Va V& 2 + 4ac 



CHAPTER II. 

INTEGRATION OF RATIONAL FRACTIONS. 

8. Preliminary Operation. If the degree of the numerator 
is equal to, or greater than, that of the denominator, the frac- 
tion should be reduced to a mixed quantity, by dividing the 
numerator by the denominator. 

For example, 

a? + l a^ + l' 

2s 8 -3a 4 + l = 2a , 3 -2^ + 3^ + 1 . 

x 4 + x 2 x 4 + x 2 

The degree of the numerator of this new fraction will be 
less than that of the denominator. Such fractions only will 
be considered in the following articles. 

9. Factors of the Denominator. A rational fraction is inte- 
grated by decomposing it into partial fractions, whose denomi- 
nators are the factors of the original denominator. 

Now it is shown by the Theory of Equations, that a poly- 
nomial of the nth degree with respect to x, may be resolved 
into n factors of the first degree, 

(x — a{) (x — a 2 ) (x — a 3 ) • • • (x — a n ) . 

These factors are real or imaginary, but the imaginary fac- 
tors will occur in pairs, of the form 

x — a + &V— 1, and x— a — &V— 1, 

whose product is (x — a) 2 + b 2 , a real factor of the second 
degree. 



RATIONAL FRACTIONS. 189 

It follows, then, that any polynomial may be resolved into 
real factors of the first or second degree, and only such factors 
will be considered in the denominators of fractions. 

There are four cases to be considered. 

First. Where the denominator contains factors of the first 
degree only, each of which occurs but once. 

Second. Where the denominator contains factors of the first 
degree only, some of which are repeated. 

Third. Where the denominator contains factors of the sec- 
ond degree, each of which occurs but once. 

Fourth. Where the denominator contains factors of the sec- 
ond degree, some of which are repeated. 

10. Case I. Factors of the denominator all of the first degree, 
and none repeated. 

The given fraction may be decomposed into partial frac- 
tions, as shown by the following example, 



J ,3-4, dX ' \ 



Assume J , 3 — 4x 

3^+6,-8 _ x 2_^_q x _s = A B C (1) 

tf-lx x(x-2)(x + 2) x x-2 x + 2' ^ 

where A, B, C, are unknown constants. 
Clearing (1) of fractions, 

a? : +6x-8=A(x-2)(x+2)+Bx(x+2) + Cx(x-2) . . (2) 

= (A + B + C)x* + 2{B - C)x -4 A. . l\\ 

Equating the coefficients of like powers of x in the two 
members of the equation, according to the method of Indeter- 
minate Coefficients, we have 

A + B + C=l, 

2(B-C) = 6, 
-±A= -8; 
whence A = 2, B=l, C = - 2. 



UV 



v^ 



190 INTEGRAL CALCULUS. 

Hence ■ — ; = - -\ ; 

x 5 - ±x xx -2 x + 2' 

and f x2 t 6X A ~ S dx = 2lo % x + lo ^ - 2 ) - 21og(z + 2) 

%J X — ~fc X 

= l x\x-2) 
& {x + 2y 

A shorter method of finding A, B, C, is the following : 

If in (2) we let x = 0, B and G will disappear from the 
equation, and we shall have 

-8 = -AA, or A = 2. 
Similarly, If x = 2, 8 = 8 J3, or 5 = 1. 

If a>=-2, -16 = 8(7, or C = -2. 

EXAMPLES. 



l r < . 3 !" i ft <to = i °gK a? + 3 ) 2 ( 8 - 2 )]- 

*/ ar -j- a? — o 

g p 2 + 2x-CO S 'a fc = a , + g eCa loga; + l+COSa - 

J ic 2 + 2ic + sin 2 a 2 x-{-l — cos a 

4. f ^ = ^!_2a;+-log ^~ 1 +— log (x + 2). 

J(> 2 -l)(z+2) 2 6 °(^ + l)3^3 ^ T ; 

J je2-4a; + l 

= 2 + ^ log(a?-2-V3)- 2 ~ V ^ log(a;-2+V3) 

2V3 2V3 

= |log(o; 2 -4 a ; + l) + -^ lQ g a; ~ 2 ~ Vg ' 



RATIONAL FRACTIONS. 191 



6. " T a dx== q+ o +4^ + log- i - 
*/ or — 4a; 3 2 (c 



x + 2) 

7. f 6( a? + 3)da; =1q 
J x 5 — 5x 3 4-lx 



x*(x-2y(x + 2y 
(x — i)\x + iy 



11. Case II. Factors of the denominator all of the first degree, 
and some repeated. 

Here the method of decomposition of Case I. requires modi- 
fication. Suppose, for example, we have 



/ 



x 3 +l 7 
ax. 



x(x — 1) ; 



If we follow the method of the preceding case, we should 



x(x — l) 3 X x — 1 

But since the common denominator of the fractions in the 
second member of this equation is x(x — 1), their sum cannot 
be equal to the given fraction with the denominator^ (x — l) 3 . 
To meet this objection, we assume 

x 3 + l A B C D 



x{x — l) 3 x (x — 1) 3 (x — 1) 2 x — 1 
Clearing of fractions, 

x 3 + 1 = A{x - l) 3 + Bx + Cx(x - 1) + Dx(x - 1)' 
= (A + D)x 3 + (-3A+C-2D)x 2 

+ (3A + B-C+D)x-A. 

Hence A + D = 1, 

-3A+C-2D = 0, 

3A + B-C+D = 0, 

-A = l. 

Whence A = -l, B = 2, C=l, D = 2. 



192 INTEGRAL CALCULUS. 



mx, n 0T 3 +1 1,2 1 

Therefore — — = = 1 + 



x(x-iy x (x-iy (x-iy x-i 

Hence 

_ + log (^l)!. 



(x-i) 2 " a? 

EXAMPLES. 
3 , ,(^-2) 2 



1. f (*-8)cfe = _3_ + log , 

f 3<e 2 — 2 , 12o3 + 19 , i , ■ on 

2. I do3 = — — + 31og(o3 + 2). 

J (o3 + 2) 3 (a: + 2) 2 &V T ; 

3 r (3o3 + 2)c?Q3 = 403 + 3 t 03 2 

J 03(03 + 1) 3 2(o3 + l) 2 & (a; + 1) 2 

J (a^ + art 1 2 ^ 03 2 +-03 bL V ; J 



1 , s+V5 



+ -^-lo 



2) 2 4(0^-2) 8 V2 03 -V2 



6 ( p 9(W+4o3+2)c7Q3 = 2o3-5 2o3+l lo „s-2 
J <v_z_2) 3 (x-2) 2 2(a3 + l) 2 & 03+l 

^-1)*, 12s + l_ 108Q3-61 41 

J (203 2 -O3) 3 203 2 4(203-l) 2 

■log (2 a — 1). 



12. Case III. Denominator containing factors of the second, 
degree, but none repeated. 

The form of decomposition will appear from the following 
example, f 5o3 + 12^ 



T 5o3- 
J x(x 2 



(^ + 4) 



RATIONAL FRACTIONS. 193 

We assume &x + 12 = A Bx + C ..... (1) 

x(x 2 + £) x x 2 + 4 v J 

and in general for every partial fraction in this case, whose 
denominator is of the second degree, we must assume a numer- 
rator of the form Bx + C. 

For it is evident that each additional fraction of this kind 
increases by two the degree of the equation, when cleared of 
fractions, and consequently increases by two the number of 
equations for determining A, B, C, •••. 

Hence its numerator should add two to the number of these 
unknown quantities. 

Clearing (1) of fractions, 

5 a; + 12 = (A + B)x 2 + Cx + 4 A. 

A + B = 0, C=5, 4.4 = 12. 

Whence A = 3, B = -3, 0=5; 

, - 5a; + 12 3 , — 3cc + 5 

therefore - 1 =-H = 

a;(ar + 4) x x- + 4 

3 a; + 5 , Q /" a;cfcc K f dx 



f=m i -- s m i +s f 



Hence 



ar^ + 4 
= -|log(^ + 4) + |tan-^ 

C 5x + 12 dx = 31og g + Stan-*. 
Ja;(ar + 4) S Vf + 4 2 2 

Take for another example, 

r (2x 2 -3x-3)dx 
J {x-l)(x 2 -2x + 5) 

This fraction is decomposed as follows : 

2a 2 -3a- 3 1 3.u -2 



(x — 1) (a; 2 — 2a; + 5) x — 1 x 2 — 2a; + 5 



194 INTEGRAL CALCULUS. 

dx 



r (Sx-2)dx = r (3x-3)dx C 
J x 2 -2x + 5~J x 2 -2x + 5 J 



x 2 -2x + 5 



= |log(^ 2 -2a; + 5) + ltaii- 1 ^:. 



f (2x 2 -3x-3)dx =1 (s*-2aH-5)* l tan -is-l 
J ( x -l)(x 2 -2x+5) & x-1 2 2 



EXAMPLES. 



- Cx 3 — 1, . l n x 2 + 3 /o. 1 x 

J x* + 3x 6 x 2 -y/§ 

o C dx 1, X 4 1, _i 

2. I — = -loar tan 1 x. 

J (a; 2 + l)0 2 + z) 4 5 (x + l) 2 (a^ + l) 2 

3. f ***" l_ + lio g (^- 1 ) 2 . 

J 0-l) 2 (> 2 + l) 2(aj-l) 4 & a^ + l 

f (« 3 -6)f^ , a: 2 + 4 ,3, _ r x 3 , _-l x 

4. I -7 J- = log — ^ + -tan x ■ tan 2 ■• 

Jd* + 6*» + 8 6 V^+2 2 2 V2 V2 

K C dx 1 -, # — 1 1, _-, 

0. I — - = -log - — -tan l x. 

J x'-l 4 5 cc + l 2 



+ 

(5a? — l)dx 



(x 2 + 3)(x 2 -2x + 5) 

log ^~ o 2x + 5 + ^tan" 1 ^— t - — tan- 1 -^-. 
x 2 + 3 ^2 2 V3 V3 



J x 2 (2x 2 -2x+o) x 2 6 2x 2 -2x+5 3 

J x 3 + l 6 V-oj + I V3 V3 

9. fi^ 



i2aj— 1 



V3 V3 

\x 2 -5)dx 

6a 2 + 25 

2 *V + 2z + 5 4^ 2 2 j 

= -log „ h-tan l -« 

2 ^ , x 2 + 2x + q 4 5-x 2 



RATIONAL FRACTIONS. 195 



10. CJ^L = JLjog^ + W2 + 1 + y^-ij^, 

Ja 4 +1 V2 x 2 -xV2 + l 1-^ 






+ 

# 2 cos2a + 1 



dx 



2a 2 cos2a + l 

sina-, o , a; 2 + 2a;sina + 1 cosa , _ 1 2a;cosa 

~4~ ° g a 2 -2asina-+-l 2 ^ 1-a 2 ' 



13. Case IV. Denominator containing factoids of the second 
degree, some of which are repeated. 

This case bears the same relation to Case III., that Case II. 
bears to Case I., and requires a similar modification of the 
partial fractions. 

Tor illustration take 

•2 a 3 + x + 3 



p 



dx. 



(x* + iy 

We assume 

2a?+x + 3 == Ax + B Cx + D 
(a 2 -f I) 2 (x 2 +l) 2 x 2 +l ' 

2x 3 + x + 3 = Cx 3 -f- Dx 2 + (A + C)x + B + D. 
A = -l, B = 3, (7=2, D = 0. 

Therefore M±^±l = ^±A + _^_. 
(^H-l) 2 (> 2 +l) 2 z 2 +l 

/-^_+3, (*xdx of da; 

(^Tl)" 2 ~J (?+l? J (S+l? 



&+•/■ 



2(x 2 +l) J (x 2 +l) 2 

To integrate the last fraction, we use the following formula 
of reduction, 

r dx = 1 |" a? , 9 qv f rf.r "| 

J (cc 2 +a 2 ) M 2(?i-l)a 2 L(a; 2 +a 2 ;r- 1 ™ " V (a^+a 2 )"- 1 ^ 



196 INTEGRAL CALCULUS. 

This formula will be derived in Chapter IV., but the 
student can now verify it by differentiating both members. 
It enables us to integrate the expression peculiar to 

/dx f* dx 
— — , by making it depend upon I . 

By successive applications the given integral is made to depend 

(* dx 1 x 

ultimately upon f 9 , which is -tan -1 -- 

To apply this formula to i ' , we make a=l and n=2. 
J \Xr -\- 1) 



We then have 



+ §tan-i* 5 



/ dx _ i r x r dx 
(aj*+ 1)2 ~ 2 [rf+ 1 + J aj»+ 1 

, C-x + 3 7 1 3x , 3. ! 

whence I — - — ! — dx = - — - — - — \- - — - — — -f -tan -1 a;, 

J (ar>+l) 2 - 2(x 2 +l) 2(> 2 +l) 2 

, r2x?+x + 3? 3a; + 1 ,3, _, (1 . , , iX 

and J ^fel^ = 2(^FT) + 2 tan 8+log (x +1) - 

As another example in the integration of a partial fraction 
in Case IV., consider 

3x }dx 



r 3x + 2 dT = f_V___2y_ 13 r 

J (x 2 -3x + 3) 2 X J (x 2 -3x + 3) 2+ 2 J " 



( £C 2_3 x + 3)2 J(^-3^ + 3) 2 ' 2J (ar*-3a; + 3) 2 

3a;-- W 
r > \ 2 J _3 / » (2x-3)dx 3 

J ^2_ 3a . + 3 ^-2j ( a j2_ 3a , + 3^- 2 (> 2 -3a; + 3)' 



/ da; _ / " cfa _ C 

( ^_3^ + 3) 2 -Jr/ _3V + 3T-J" 



dz 



[(-1MJ •'MJ 



3 

where 2 = x 

2 



RATIONAL FRACTIONS. 197 

Applying the formula of reduction, 

dz 2/ z r dz \ 2 z 4 ,2z 

:tan _1 - 






3 

Or substituting 2 = a; , 

z 

C *° = 2 *~ 3 + JLtan-^Z^ 

J(3*_3a; + 3) a 3(^-3^ + 3)^ 3 V3 V3 

hence 

13(2a?-3) 



r (3a; + 2)dx = 3 

J (^-3» + 3) 2 2(a? 2 -3x-f 



3)"* 6(^ 2 -3«-f-3) 

. 26 , _j2aj-3 
-I -tan l — 

3V3 V3 



13x-24 , 26 , ^x-3 

= —7-: - +■ tan x 

30 2 -3a + 3) 3V3 V3 



EXAMPLES. 



p3 +a ._ 1 ^~ 2 + J:iog <y + 2 ) 5-tan- 1 -^-. 

J (z 2 +2) 2 4(z 2 +2)^2 8V ^ ; 4V2 V2 

' /V-<e 4 -f 21, 4cc-9 , 3jc + 6 , 1, .« , Q \ 

' J^4f- & = 4^3)' + 2(^ 3 ) + 2 10 ^ +3) 



1 1 j. _i SB 

H tan * — -« 

2V3 V3 



o /V-2a-j-l , 3x 2 +x-{-2 , , (V-hlV* 3, , 

3. I ^—dx— - — — — hloe^ — — — ' tan -1 a. 

J x\x 2 +l) 2 i 2x{x i +l) & a; 2 2 

r (4;x 2 -8x)dx 3x 2 -x ., (a? -IV , , , 

4. I — ^ '- =— — — — -hlos;^ ^-htan -1 #. 

J (x-lfix'+lf (x-l){x' + l) & ar+1 



198 INTEGRAL CALCULUS. 



5 rJ+sx+n^ 3(*-7) + i log(x ,_ Ax+9) 



§^tan-^^. 



J (a 4 + ar 



V5 

+ 1) 2 3(> 4 +a 2 +l) ' ~~ l V+a; + l 

+ Aftan-^±l-ltan-2£^ 
V3V V3 3 V3 



l)(to_ 2(s 2 -l)(s-l) |log ^-a; + l 



CHAPTER III. 

INTEGRATION BY RATIONALIZATION. INTEGRATION 
BY SUBSTITUTION. 

14. As the preceding chapter provides for the integration 
of rational fractions, it follows that any rational algebraic 
function is integrable. 

Some irrational expressions may be integrated by substitut- 
ing a new variable, so related to the old, that the new expres- 
sion shall be rational. 

15. Expressions involving only fractional powers of x. Such 
forms may be rationalized by assuming x = z n , where n is the 
least common multiple of the denominators of the several 
fractional exponents. 

Take for example, J x . 

J x% + x^ 

Assume x = z% dx — 6 z 5 dz ; 

then x? = 



z 3 , & 



/' dx _ r6z 5 dz _ c f 
™i i ~k J z 3 + z 2 J j 



z 3 dz 



X? _|_ X S ^ * " + Z J Z + 1 



dz 



= |-| + *-log(* + ±). 



Substituting in this, z = X®, we have 

f t dx = 2x% - 3x* + 6a>*- 61og(a?* + 1). 

J or? J- ^3 



X 2 +X< 



200 INTEGRAL CALCULUS. 

16. Expressions involving only fractional powers of (a -f- bx), 
may be rationalized by the method of the preceding article. 

dx 



Take for example, I - 



(x _ 2)* + (a? - 2)* 

Assume x — 2 = z 6 , dx = 6 z 5 d2. 

J (x-2f + (x-2f~* z 5 + z*~ Jz + 1 
= 6[z-i>g(* + l)]. 

Substituting 3 = (as — 2) ¥ , we have 



'« 



da; 



a;-2)°+(a;-2)< 

EXAMPLES. 



= 6(x - 2)* - 6 log[ (a? - 2)* + 1]. 



L r_^ = ^i^4 log(a;f+1)> 

2. f_J5 ^-4 + iog (^ + 1 ) 6 . 

J x^ -f x% x* x 

3. f?l±±clx=-^+^- + 2logx-24:log(x^ + l). 

Jx^ + x^ X 6 X 12 

4. C dx == §^ + 21o g ^l + 4tan- 1 ^. 

J x* — x^ 3 X s + 1 






J xVx + 1 V a> + 1 + 1 

6 / » aftfcc = 6a; 2 + 6a; + l 
J (4 a? + 1)* 12(4a; + l)^ 

7. f ^ == = ?(o; + l)*-3(aj + l)*+31og(l + ^ : f : I). 

•^l+Vaj + l z 



INTEGRATION BY RATIONALIZATION, 201 

8. 






-1)% 



9 r(2s-l)«d 
J a>-(2s-l 



(2 « -If 



= 4(2.-l)* L - 2(2^-1)* +31og (2,-l)^-l 
(2a-l)*-l (2»-l)* + l 



* *» cc ctcc 



iCCti 

(2 cc + 2)4 +4(2^ + 2)* 

=|(a + 36)(2z + 2f-|(2x+2)4-28tan- 1 ^ 



17. Expressions of the form f(x 2 )>xdx, involving fractional 
powers of (a + bx 2 ), may also be rationalized by the method 
of Art. 15. 

Take for example, I • 

J -Vl-x 2 

Assume 1 — x* = z 2 , x 2 = l — z 2 , xdx = — zdz. 
J Vl-x 2 J z J 



"~| 3 ) = ~l (3 ~" 2) = ~ 3 Z ^ ( ^ + 2) * 



EXAMPLES. 

- C rfdx 3a 4 — 2^ + 2 / o 2 , -. 

1. I = . V2a; 2 + 1. 

^V2x 2 + 1 30 

2. fz> 2 - x 2 )klx = -5-(6x* - a 2 x 2 - 5a*) (a 2 - x 2 )*. 

3 . r__g = _ jL 1ng vg+ « 2 - « = lio g g — 

JaVar' + a 2 2a -vV + a' + a a V» 2 + a 2 + 



202 INTEGRAL CALCULUS. 

4 



. f 3/ _^ = §p^ + (^+i)*+io g (-^+i-i)l 

J Vr+1- 1 ^L ^ J 

. f Xdx _ =^log(V3^^ + l) + jlog(V3^^-3). 

J a? + 2^/3-0* 4 4 



18. Expressions containing V x 2 + ax -J- 6. 
If we assume, as in the preceding articles, 

V^M-ax+T = 2, x 2 + ax -f- b = z 2 , 

the expression for x, and consequently that of dx, in terms of 
z, will involve radicals. To meet this objection we assume 



Vx 2 + ax -f b = z — x, ax + 6 = z 2 — 2zx, 

^-^ ^ = 2(z* + az + 6)cfe 

2z+a (2z + a) 2 ' 

/-o— ;— r z 2 + az + 6 

Vx 2 -f ax + o = z — x = — ^ -£ — 

2z + a 



Thus Vx 2 + ax + 6, x, and dx are expressed rationally in 
terms of z. 

dx 



Take for exam 



pie, P 



: Vx 2 — X + 2 



Assume V x 2 — x-f-2 = z — x, — x + 2 = z 2 — 2zx, 

g - *- 2 3 . 2(z 2 -z + 2)dz 

a? -2^1 ? * (2z-l) 2 ' 



Vx 2 — x + 2 = z — x = 



z 2 - z 4- 2 



2z-l 

/ dx _ r 2dz 1 , z— V2 

xVx 2 -x + 2 Jz 2 -2"V2 g z+V2 



Substituting z = Vx 2 — x + 2 -f x, 

dx 1 , Vx 2 — x + 2-fx — V2 



r — ^ — =^io g 

J W.x 2 — x 4- 2 a/2 



Vx 2 - x + 2 V2 Vx 2 -x + 2+x + V2 



INTEGBATION BY RATIONALIZATION. 203 



19. Expressions containing V— x 2 + ax + b. 

To rationalize in this case, it is necessary to resolve b-\-ax— x 2 
into two factors. These factors will be real, unless the given 
radical V& + ax — x 2 is imaginary for all values of x. For 

b + ax-tf^ + b-fe-xj 

= |~1 ( V^+46+a) -ajT| ( Va^+Ib-a) +xl. 

These factors are real unless a 2 + 4 b is negative, but then 
b + ax — x 2 is negative for all values of x, and consequently 
V& + ax — x 2 is imaginary. 

Eepresent the two factors thus, — 

b 4- ax — a 2 = (a — x) (/2 -f x). 
Now assume 



V& + ax — x 2 = V(a — x) (ft + x) = (a — x)z 

then B +x == (a — x)z 2 , x = a ~" f^ 

Thus x is expressed rationally in terms of z. 

dx 



Take for exam 



pie, j - 



x V2 -f- x — x 2 
Assume V2 + x — x 2 = V(2 — x) (1 + x) = (2 — x)z. 



2s 2 -1 _ 7 „ 62dfe 



1 -j-a; = (2— x)z 2 , x = — — — , dx = 



z 2 + l' (z 2 + iy 



V2 + x - x 2 = (2 - 0)0 = 



z 2 + l 
Therefore, 

/ dx __ r 2dz _ 1 1q gV2~l 

0JV2 + X-X 2 J 2z 2 ~l V2 ° & zV2 + l 



204 INTEGRAL CALCULUS. 



Substituting z = -J + x f 

* Zi — X 



clx 1 , V2 + 2«-V2- 






;c 



^V2 + ^-^ V2 V2 + 2a?+V2-a; 



EXAM PLES. 



1. f == ^ = =2tan- 1 (« + V^ + 2^-l). 

2 r acta _ 8 + 6a; 

J (2+3a-2z 2 )f~25V2 + 3a-2ar 2 



J da: 
a^s/x 2 



cto - 1 Vx 2 - 2 

= ■ or = 



2 x{x+Vx*-2) 



2x 



r Vx> + 2x dx = i + log(x + l + Vrf+2x) 

J x 2 x+Vx* + 2x 



or 



2\^i^ + 21og(Vz + 2+Va). 



5 jvra d __ 2 ^ +2tan _^ 



6-a 



x 






= -2^ + eos-^. 



c7» 



(aj-l) 2 V^-2x + 2 
1 



vV-2a; + 2 + a; Var 2 - 2a + 2 + z -2 



a/^^2^ + 2 

or =--^ -- 1 — . 

a; — 1 



20. Integration by Substitution. This method is used for 
rationalization, as shown in the preceding articles, but in other 
cases the introduction of a new variable often simplifies 



INTEGRATION BY SUBSTITUTION. 205 

the given expression, and renders it directly integrable. This 
is illustrated by the following examples. 

EXAMPLES. 

C a?dx 18a 2 + 27 a + 11 , , , . iN 
2 - J(^l)i = 6 (* + !)■ + l0 ^ + 1 >- 

Assume x + 1 = z. 

dx 1-, x A a 



3 # I — = -log— — Assume x = - 

J x Va 2 + or a a -f Va 2 + x 2 z 

4. I =-log x Assume x = -- 

J x^/a 2 — x 2 a a-\- Va 2 — x 2 z 

5. f x * dx , = § (x 2 - 3) (x 2 + l)i Assume x* + l = z. 
J (x 2 + l) t 8 

O. I - — - = (# + a)C0Sa — smalogsin(a3 + a). 

•/ sm (05 + a) 

Assume x-\-a = z. 
7. f e ^ t = A (3e* - 4) (e* + 1)* Assume e* + l = z. 

8 * J*5=S? = 27 ~ I + l log (e * ~ 2) ' Assume e = z ' 



Assume x4-~ = 
^ x 

10. f (l+g* , )**» 
J x 

= ^(l+2^)t + 2tan-Xl-K2^)^+log( 1+2 ^ i - r 

n ly (i+2a> n )*+ij 



Assume 1 + 2 x n = 2 4 . 



CHAPTER IV. 

INTEGRATION BY PARTS. INTEGRATION BY SUC- 
CESSIVE REDUCTION. 

21. Integration by Parts. From the equation 
d(uv) = udv + vdu, 
we obtain, by integrating both, members, 
uv = I u dv + I v du. 

Hence I udv — uv — I vdu (1) 

The use of (1) is called integration by parts. 
Let us apply it, for example, to 





f xlogxdx. 










Let 


u = logx, 


then 


dv = 


- xdx; 




whence 


, dx 
du-= — , 

X 


and 


v = 


X 2 

2* 




Substituting in (1), we 


have 










i logX'Xdx 


= logx 


X 2 


rx 2 

J 2' 


dx 

X 






X 2 , 

= 2 l0£ 


[X - 


X 2 . 

"4* 





(2) 



The student should carefully notice how the factors u, dv, 
v, du, occur in the process, so as to be able to apply it without 
such a formal substitution as in the preceding example. 

On referring to the equation (2), we see that, after selecting 
for u a certain factor of the given integral, as log a;, we obtain 
the first term in the second member, by integrating as if this 



INTEGRATION BY PARTS. 207 

factor were constant j also that the expression following the 

second J , is the same as the preceding term, with the factor 

logx replaced by its differential. 
Take for another example 

I xcosxdx. 

Assuming u = cos x, we find 

/x 2 rx 2 
xvo$xdx = coscc I —(—since die). 

But as the new integral is no simpler than the given one, 
we gain nothing by this application of the process. 
If, however, we let u = x, we find 

I xcosxdx = xsinx — I smxdx 
= x sin x -f- cos x. 

EXAMPLES. 

1. Jx*logxdx = ^flogx — \ 

2. I x n ~ 1 logxdx = — (logx V 

J n\ n) 

3. j xsu±xdx = — xcoscc-f since. 

4. Cxlog(x + 2)dx = (x 2 - 4) log Va + 2 - - + x. 

\j 5. Cxe ax dx = —(x-~\ 
J a \ a J 

6. Cxtasi^xdx = ^-tltan^a - -. 
J 2 2 

7. i sm- 1 xdx = xsm~ 1 x + \ / l — x 2 . 

C x 2 

8. I cctan-cccfcc = cctancc f-logcosce. 



208 INTEGBAL CALCULUS. 

9. r i°g(* + l)^ = 2 V^Tl[log(* + l)-2]. 
J Vx + 1 

10 r io g (i +yx)dx = 2(1 + v - )log(1 + v - } _ 2V - 

11. I tan _1 V^^ = (l + «)tan _1 V^ — V#. 

13. I a? 2 sin _1 aj da? = — sin _1 aH — - - Vl — x?. 

«/ 3 9 

In each of the following examples integration by parts must 
be applied successively. 

14. Cx 2 e x dx = (x 2 -2x + 2) e*. 

15. f^dx=(xS-^ + 6 -*-«-Y-. 
J \ a a 2 or J a 

16. JV(log«0 2 ^ = |T(loga0 2 - \logx + 11. 

J x* 3a^L 6 J J 

. 

22. Formulas of Reduction. These are formula by which 
the integral, „ 

J x m (a + bx n ) p dx, 

may be made to depend upon a similar integral with either m 
or p numerically diminished. There are four such formulae, 
as follows, — 

f x m (a + bx n ) 

n.— n+] 

(np 



dx 
+m+l)b (np+m+l)bJ X ( a+bx > <**> ( A > 



FOBMJJLM OF B EDUCTION. 209 

Cx m (a -\-bx n ) p dx 

wp+m + 1 npH-m+1*/ 

(m + l)a (m+l)a J (+ J ' (t,) 

J ^(os + ftaf^da; 

ra(p + l)a n(p+l)a J s J K ; 

Formula (^1) changes m into m — n. 
Formula (B) changes p into p — 1. 
Formula (C) changes m into m + n. 
Formula (Z>) changes p into p -f 1. 

Formulae (C) and (D) are used when m or p is negative, 
requiring an algebraic increase. 

23. Derivation of Formulae (A) and (C). Let us put for 

convenience 

z = a -f 5af, cfe = nbx n ~ l dx. 

Then faf(a + ^ n ) p cte = Cx m z p dx = fa^+Vaf- 1 ^. 
Integrating by parts, with w = x m ~ n+ \ 

Cx m z p dx=x m ~ n + 1 — — m-n+l f^-n^+i^. 

J nb(p+l) nb(p + l)J 

nb(p+l) fx m z p dx = x m - n+l z p+1 —(m—n + l) Cx m - n z p+1 dx. (1) 
Cx m - n z p+l dx= C(a + bx n )x m ~ n z p dx=a Cx m ~ n z p dx+b Cx m z p dx. (2) 



210 INTEGRAL CALCULUS, 

Substituting (2) in (1), and transposing, we have 

(np+m + 1)5 Cx m z p dx 

=x m - n+1 z p+1 — (ra — Ti + 1) a Cx m ~ n z p dx. ... (3) 

Dividing by (np + ra + 1)6, we have (A) . 
If in (3) we substitute 

ra — n = ra', m = m 1 -\-n, 

and transpose, we have 

(m' + l)a Cx m 'z p dx 

= x m '+ V +1 — (np + m' + n + 1) 6 Cx m ' +n z p dx. 
Omitting the accents, and dividing by (ra + l)<x, we have (C). 

24. Derivation of Formulce (B) and (D). If we integrate 
by parts I x m z p dx, calling u = z*, we have 

fx m z p dx = z p -^ ^_ f^+V-V" 1 ^. 

J ra + 1 ra + 1./ 

(m + 1) Cx m z p dx = x m+1 z p — nbp J rt n + n z p - 1 dx. ... (1) 
J x m z p dx = J (a + bx n ) x m z p ~ l dx 

= a Cx m z p ~ l dx + b Cx m+n z p - l dx. . . . . (2) 

Eliminating from (1) and (2), J cc m+n 2 p_1 c^, we have 

(np + m + 1) Cx m z p dx = x m+ V + wpa Cx^^dx. . . (3) 
Dividing by rip + m + 1, we have (B)« 



FORMULA OF REDUCTION. 211 

If in (3) we substitute 

p-l=p', p=p' + l, 
and transpose, we have * 

n(p f +l)a Cx m z p 'dx = — x m+1 z p,+1 + (np'+n+m+l) Cx m z p ' +1 dx. 

Omitting the accents, and dividing by n(p+l)a, we have (D). 

Formulae (A) and (B) fail, when np -\- m + 1 = 0. 
Formula (C) fails, when m -f 1 = 0. 

Formula (D) fails, when p + 1 = 0. 

EXAM PLES. 

11 X dX X / o •> . Cki • ^_-\X 

. I — = ^Jar — x--\ — sm -• 

J Va 2 - a 2 2 2 a 

Here f = = Cx 2 (a 2 -x 2 YMx. 
J Va 2 — ar 2 *^ 

Apply (A), making 

ra = 2, % = 2, p = — -, a = a 2 , 6 = — 1. 

Z 

Cx>(a? - aj 2 )-^ = X ( a2 -^Y _ _^!_ f( a 2 _ X 2yi dx 

= _^(a 2 -^ + ^sin- 1 -. 
2 V y 2 a 

2. f Va 2 + a?dx = ^ Va 2 + x 2 + flog(a + Va 2 + a 2 ). 
*/ z z 

Apply (JB), making 

m = 0, ?i = 2, i> = -, a = a 2 , 6 = 1. 

Z 

C(d> + tf)klx = X {a 2 + x 2 )?+^C dx 

J Z ZJ (tf + y?)$ 

= | (a 2 + x 2 ) * + flog (» + V^T^). 



3. f dx = - Va ' 2 -^4-J-log- 

•J xVa 2 -^ 2 2aV 2 a 3 a 



4- Va 2 — ar 2 



212 



INTEGRAL CALCULUS. 



Apply (O), making 

m = — 3, n = 2, p = — -, a = a 2 , b = — l. 

* 2 

fx- s (a 2 - a*y*dx = x ~ 2 ( a2 ~f) 2 L_ f x -i(a a - xtykx. 

— .Z tt — A Qi */ 

Ex. 4, p. 205, gives 

fx-\a 2 - x 2 )~hx = f — = hog 

J J x-\/a 2 — x 2 a a + Va 2 — x 2 

Substituting, we obtain the complete integral. 

4 r da; = (3a 2 -2x 2 )x 
J (o«-af)t 3a 4 (a 2 -a; 2 )' 

Apply (Z>), making 

5 
m = 0, n = 2, p = , a = a 2 , 6 = — 1. 

Li 

Ex. 33, p. 180, gives 
C(a 2 -x 2 y*dx= C 



dx 



(a 2 — a? 2 ) 2 a 2 s/d 
Substituting this, we have 

2x 



/ dx _ 
(a 2 -a 2 )* 3a 2 ( 



+ 



a 2_ (B 2)f 3 a 4(a s — a?)* 



14 



2( 



3a 2 (a 2 -^ 

a; 3a 8 -2a 8 

3a 2 (a 2 -^ a * 



a 2 J 



/ 



x 2 dx x 



-\/x 2 + a 2 log (a; + Va,- 2 -f- ^= 

VaT^H? 2 2 



FORMULA OF REDUCTION. 213 

%/ 8 8 

7. I V a 2 — x 2 cfa = % Va 2 — x 2 -f ^- sin -1 — 
J 2 2 a 

8. (\fezi5da; = V(a -»)(& + ») + (a + ^sin" 1 -J^+A 
«y \6 + x \a + & 

Substitute 6 + a; = z 2 ? and the integral takes the form of 
Ex. 7. 



r\x±a { 

Jylx + b 



ax 



= V(» + a) (x -f 6) + (a — 6)log( Va; + a + Va; -f 6). 

Substitute cc + 6 = z 2 , and the integral takes the form 
of Ex. 2. 



10. C-V2ax-x 2 dx = - — - V2 aa; - x 2 + -vers" 1 

I V2 ace — a; 2 c?a; = j Va 2 — (a? — a) 2 dx, 
which is in the form of Ex. 7. 



x 



11. fx 2 Va 2 - aftfo = - (2 a; 2 - a 2 ) Va 2 - a 2 + -sin" 1 -. 
»/ 8 8 a 

12. fa; 2 V^M^aa = -(2a^+a 2 )Va^^ 

*/ 8 8 

13. f (a 2 - oj 2 )*^ = |(5 a 2 - 2 a; 2 ) Va^^ + 3a 4 sin -i* 
•/ 8 S a 

14 f __^L__ = (15 a 4 - 20 aV -f8 a; 4 ) a; 
J (a 2 -a; 2 )^ 15 a 6 (a 2 - x 2 )? 

15. f dx 

J (x 2 + a 2 ) n 

= 2(n-l)a 2 L(a; 2 + a 2 ) n - 1 + (2n ~ 3) J (rf + c^^J" 



214 INTEGRAL CALCULUS. 



Jx\x 2 + 2) 6^^4 iC ^4V2 V2 

i7. r_^_ = _ 2 (3a , 6 +4iC 3 +8) Vl^^. 

J yi _ ^ 45 

18. f ^ = - 2a * + 1 yT=^. 

19. f xdx = - ■y/2ax-x 2 + avers- 1 -. 
Here f gd * = f_^ 



a; 
Apply (A), and the integral is reduced to 



/ 



dx _,x 
= vers -• 



■\l2ax-x 2 <*> 



x^/2 ax — x 2 ax 



20 C dx ~^ 2 ax — x 2 

21 f—?^L- — a; OT - 1 V2ax— a; 2 (2m-l)<x r_of^dx_ 
J -Wtax — x 2 m m J V2 



V^az-x 2 ™ m J ■y/2ax-x 2 



22. J- 



x m -\2 ax — x 2 

-f— 



V2aa: — a; 2 , ra — 1 /" da; 



(2 m — l)aaj m (2 m — l)aJ x m ~W2ax — x 2 
23. Cx m -V2ax — x 2 dx 



x m- 



H2ax-*)* (2m+l)« (VvzUS^Sfe. 
m + 2 m+2 J 



24 



/ 



V 2 aa? — a^efa; 



a; 11 



(2 as -a 2 )* m-3 r V2aa; - aftfog 

(2m-3)aaj m (2m-3)aJ ~ a; 1 "- 1 



25. 



CHAPTER V. 

TRIGONOMETRIC INTEGRALS. 

Required j tan n sccfoc, or j cot n xdx. 



These forms can be readily integrated when n is an integer, 
positive or negative. 

f ta,n n xdx = j tan n_2 ic(sec 2 cc — l)dx 

= J t&n n ~ 2 xsec 2 xdx — J t3Ln. n ~ 2 xdx 

tan n_1 sc Cx. n-2 ^ 

= I tan n z x dx. 

n — 1 J 

Thus j taxfxdx is made to depend upon f tSLn n ~ 2 xdx, and 

ultimately, by successive reductions, upon j tanxdx or I dx. 
When n is negative, the integral takes the form 

j cot w a? dx, 

which can be integrated in a similar manner. 
For example, required J t&tfxdx. 

J t&tfxdx = j tan 3 a;(sec 2 x — l)dx 

tan 4 a? C. «, , 

= I tan^cc dx. 

4 J 

I taxfixdx = | tan#(sec 2 £ — l)dx 

tan 2 # t 
= lo^seca;. 

2 & 

Hence I tan 5 x dx = — — -f- log sec x. 

*/ 4 £ 



216 INTEGRAL CALCULUS. 

26. Required J se,o, n xdx, or J cosec w # dx. 

These forms can be readily integrated, when n is an even 
positive integer. 

I sec n xdx = I sec n_2 a;sec 2 iccZa; 

/ n-2 
(tan 2 # + 1) 2 sec 2 #c7a?. 

If 7i is even, will be a whole number, and the first 

' 2 

factor can be expanded by the Binomial Theorem, and the 
terms integrated directly. 

The following example will illustrate the process. 

I sec G xdx= I sec 4 xsec 2 xdx 

= J (tan 2 a; + 1) 2 sec 2 xdx = J (tan 4 # + 2 tan 2 # -f- 1) sec 2 cc dx 

tan 5 # . 2tan 3 aj . , 

= — — H _+tana. 

o o 



27. 



Required J tan m icsec n #c?#, or J cot m #cosec M #cfa;. 



These forms may be readily integrated when w is a positive 
even number, or when m is a positive odd number. 
When n is even, the method of Art. 26 is applicable. 
This is illustrated by the following example : 

I tan 6 ajsec 4 ajc?a; = j tan 6 #(tan 2 £ + l)sec 2 xdx 

= I (tan 8 # -f- tan 6 »)sec 2 ccd. / c = — - — | — ■ — • 
When m is odd, proceed as follows : 
I tan TO #sec n £ecfcc= I tan m_1 £csec n_1 a;seca;tana:c?x 

/m— 1 
(sec 2 ic — 1) 2 sec n-1 #sec#tana;cfa;. 



TBIGONOMETBIC INTEGBALS. 217 

Since m is odd, m ~ will be a whole number, and the first 
factor can then be expanded, and the terms integrated 
separately. 

The following example illustrates the process. 

J tan 5 a; sec 3 # dx= I tan 4 # sec 2 # sec x tan x dx 

— I (sec 2 # — l) 2 sec 2 icsecictanicc?aj 

= J (sec 6 x — 2sec 4 sc + sec 2 sc)seca;tana;cfas 

_ sec 7 a; _ 2 sec 5 a , sec 3 a 

EXAMPLES. 
1. ( tan 3 acfa? = _^E_^ _j_ log cos x. 

2/ i « t xian x xan 3? , ■ 
I tan b a?aa; = — — + tan x — x. 

J 5 3 

3. | cot 4 -dx = — cot 3 - + 3 cot- -f- x. 
J 3 3 3 

4. ftan 5 -cfa = tan 4 - - 2 tan 2 - + logsec 4 -. 
J 4 4 4 4 

K /• ,- 7 cot 6 # . cot 4 a; cot 2 a? •, 

0. I cot 7 icc?cc = — j — logsmjw. 

•J o . 4 2i 

6. When n is even, 

/, _ 7 tan n-1 a; tan w-3 # , tan n_6 a; 
t&n n xdx = ■ J . 
71 — 1 71—3 71 — 5 

-f(-l) 2 (tana — a;). 



218 INTEGRAL CALCULUS. 

7. When n is odd, 

J n — 1 n — 3 



?i — l n — 3 ?i — 5 

1*4-1 



+ (-l) 2 (^-logsecA 

8. f sec 8 z <fc = t -^ + ^^ + tan 3 z + tan x. 
J 7 5 

o /* 6f) , cot 5 2a; cot 3 2# cot2# 

9. I cosec 6 2 x dx = — — 

u 10 o u 

a A 4 4 j tan 7 a? . tan 5 aj 

0. I tan 4 aj sec 4 # ax = -\ — - • 

J 7 5 

/ 

o /*. s . 2tan% , 

2. I tan 2 £ sec 4 £ cZ# = ■ -f- 

J 5 9 



i sec G xdx , o . cot 3 cc 

1. I- = tan# — 2 cot a; 

tan 4 # 3 

2tan-# , 2tan 2 £ 



/• , 5 4 , cot 6 & cot 8 # 

3. I cot 5 # cosec 4 ^ ac = 

J 6 8 



4. I tan 8 a? sec 5 z dx = — - ^- 



/• 



k r 4.5 57 cosec 9 # . 
5. I cot 5 3 cosec D 2 dx = — ■ — f- 



cosec 9 # . 2cosec 7 x cosec 5 oj 



5 



6. I tan 5 #sec 2 #d£ = 2sec 2 a? — — \- 

7. J (tan a; -f- cot x) 3 dx = - (tan 2 # — cot 2 #) + log tan 2 #. 

o /~sec% + 1 7 tan 7 # , 2tan 5 # . 2tan 3 # . 

8. I ! — dx = h x. 

J sec 2 z + l 7 5 3 

9. I (sec x + tan £) 4 cfo = - (sec 3 # + tan 3 a) — 4 sec x -f- cc. 



TRIGONOMETRIC INTEGRALS. 219 



28. 



Required i sin OT a; cos n cc dx. 



This is readily integrated when m or n is a positive odd 
number, or when m-\-n is a negative even number. 
Suppose n to be odd and positive. 

I sin. m xcos n xdx = J sin m a?(l — snrV) 2 cosicfc 
As ^~ is a positive integer, the second factor can be 

A 

expanded, and the terms integrated separately. 
For example, 

I sin 2 #cos 5 #cZ# = I sin 2 #(l — sm 2 x) 2 cosxdx 

= I (sin 6 # — 2sin 4 # -f- sin 2 #) cos acta 
_ sin 7 a? _ 2 sin 5 # sm s x 

A similar process may be used, when m is odd and positive. 
For example, 

I sin 3 #cos 2 #cfoc = f cos 2 #(l — cos 2 x)sinxdx 

= I (cos 2 ic — cos 4 x)sinxdx 

_ cos 3 # cos 5 a; 

When m -f- n is a negative even number, the form can be 
integrated by expressing it in terms of sec a; and tan#. Thus 



/Km m xcos n xdx= r^^cos"^^ 
J cos m x 

= J tan^sec - "* - ^^. 



220 INTEGRAL CALCULUS. 

Since —m — n is positive and even, the method of Art. 27 
is applicable. 

/sin x 
- dx. 
cos 4 # 

Here m — 2, n = — 4:, m-\-n = — 2. 

/sin 2 sc , f, j o , tan 3 ic 
— — ax = I tan J #sec^a# = • 
cos 4 ic J 3 



o,o$?xdx 



EXAMPLES. 



sin 5 a? sin 7 # 



cos 5 a , 2cos 7 a? cos 9 # 



1. I sin 4 # 

o C • «, a -, COS 5 £ . 2 COS 7 ; 

2. I sm 5 # cos 4 # cto = 1 

•J o 7 

o C • i * cos 7 x 3cos 5 # . - 

. 3. I siirxdx = h cos d # — cos x. 

J 7 5 

4. | cos 5 -dec = sin 5 - sin 3 ^ + 5sin^. 

J 5 5 3 5 5 

c /"cos^cto sin 3 # . 1 

5. | — — — = 2sma> : 

*/ sura; 



sin a? 



/> /* • s 3 / ^ o 3 / /cos a; cos 3 # . cosV\ 

6. I sura; -\/cosxdx= — 3 vcosa?( • -f- < 

J \ 4 5 16 / 

» rcos^d^ _ _ cot 5 a; 
*/ sin 6 a? 5 

C dx tan 3 # . , , 

8. 8 = h 2 tan #— cot x. 

J sm 2 a;cos 4 a; 3 

ft C dx tan 4 a* . 3tan 2 # cot 2 # . 01 , 

9. -r- 3 j- = — T" + — o~~ ^- + 31ogtana. 

J sm 3 a;cos 5 a; 2 

■r- 



2 5 

1 « i 'sin^ dx 3 tan" 5 x 



cos 3 a; 



-ii C d% 2Vtan£c /4 .„ «. \ 

11. I — - = (tancc — 3cotcc). 

J Vsin 3 iccos 5 ic 3 






TRIGONOMETRIC INTEGRALS. 221 

f* 8 

12. I (sin x -f cos x) h dx = sin x — cos x + - (sin 3 x — cos 3 cc) 

(sin 5 a? — cos 5 ic). 

5 

13. C(— — Ycfo= -(tan 3 o; - cot 3 a) - 2(tan 2 cc - cot 2 x) 

*/ \sincc cos ay 3 



14 J 



+ 7 (tan a; — cot x) — 8 log tan «. 

sin"- 3 ^. . ,., tan w+2 05 2tan w+1 05 , 3tan M x 

(sm a; cos a; — lYdx = — I 

cos n+3 x n ■+- 2 71 + 1 n 

_ 2tan w - 1 a; tan n " 2 a; 



w-2 



29. Integration by Multiple Angles. By means of the proper 
formulae of trigonometry, sin m sc cos n #, when m and n are pos- 
itive integers, can be expressed in a series of terms of the 
first degree, involving sines and cosines of multiples of x. 

If we use the method of Art. 28 for integrating terms with 
odd exponents, occurring during the process, the following 
formulae for the double angle will be sufficient for the trans* 
formation of the terms with even exponents. 

2 sin x cos x = sin 2 x, 



2 sin 2 ic = 1 — cos 2 x, 



2cos 2 z = l + cos2a;. 
For example, required 

I sin 4 05cos 2 cccfa5. 

sin 4 a5COS 2 sc = (sino5cosic) 2 sin 2 a; = - sin 2 2a5(l — cos 2 a;) 



— - sin 2 2 £C cos 2 05 + -— (1 — cos 4#), 
8 16 



Hence I sinWs^z= s{y)?2x ■ x sin4a! 



i sin 4 ; 



48 16 64 



222 INTEGRAL CALCULUS. 

EXAMPLES. 

1. f^x<te = \(^- S m2x + ™f*\. 

2. j ao ^xdz = \(^ + sm2x + ^\ 

3. i sm 2 xcos 2 xdx = -(x . 

4. Csm e xdx = ~(5x-4,sm2x-\- sm32x Jh-sm4,x\ 
J 16 V 3 4 y 

5. Ccos 6 xdx = — (5x + ±sm2x- s -}?^ + -sm4:x\ 

J i6V 3 4 y 

6. fsin 4 a;cos 4 a;^ = -^-f'3a;-sm4a; + ^- 8 ^ 
%) 128 V o 



7. | cos 6 #sin 2 o;cta = ( 5x-\--sm 3 2x — sin4a; — sm ^ V 

J 128V 3 8 j 

8. I sin 8 a;cZa; 

1 /35 a? a • n ,2.o .7.. . sin8a?\ 

= — 4sm2£c + -sm 3 2a; + -sin4»H • 

16V 8 3 8 64 y 

30. Integration of Trigonometric Functions by Transforma- 
tion into Algebraic Functions. 

If in the integral j sm m xcos n xdx, we assume sin a; = 2, we 
have also 

cos x = (1 — z 2 ) *, x = sin _1 z, dx = 



Vl 



z- 



//* - dz 

sin m a;cos n a;<2a; = ( z m (l — z 2 ) 2 — - 
J Vl-* 2 



= JV(1-Z 2 ) 



w-l 

2 dz. 



TBIGONOMETBIC INTEGBALS. 223 

By means of the formulae of reduction, this -form is inte- 
grable for all integral values of m and n, positive or negative. 

In the preceding transformation we might have assumed 
cos x = z, instead of sin x = z. 

Any expression containing sin x and cos x, free from radicals, 
can thus be integrated, either by a formula of reduction or by 
rationalization. Moreover, since the other trigonometric func- 
tions can be expressed rationally in terms of the sine and 
cosine, it follows that any rational trigonometric expression 
can be integrated. 

EXAMPLES. 

cos 3 cc cos 5 aA sin a; ■ x 



- C • 2 4 ^ /cos x . cos 3 cc cos 5 a?\sm 
1. j sm *Wa,< to= =(- i - + _-_j_ 

Assume cos x — z, sin 2 a; = 1 — z 2 , dx = 



16 

dz 



-VI -z 2 
| sm 2 xcos i xdx = — I z i (l — z 2 ydz. 

By the formulae of reduction, 

fzHl-z 2 ydz = ±(--^- z -)(l--z 2 )%--}~cos- 1 z. 
J K J 2\3 12 8/ } 16 

Substituting z = cos x, we have the integral required. 

o C <? 7 secajtana; , 1-, , . , N 

2. I sec 3 x dx = \--log(secx + t2Lnx). 

Assume secas = z, x = sec -1 2, dx = — • 

zVz 2 — 1 

fsec*xdx = f z ' dz =g-V^Ti + llog( g+ V^Zl) 
J J ~\/z 2 — 1 2 2 



■f 



sec x tan a? . 1-. / . , x 

= h -log(seca; + tana;). 

dx 1 . -, , x 

+ log tan - 



since cos 2 a; cosx 2 



Assume tan x = z. 
x 
2 



224 INTEGRAL CALCULUS. 

4 - f ■ ."" a -=^f— J^ + |logCeQ» + tan»). 

c/ sm 2 a;cos 3 a; 2 cos 2 a; sin a; 2 

K /*cos 4 #cfa cos a; 3 n a; 

0. I — — — = — — — cos a; --log tan-- 

J sin 3 a; 2 sin 2 a? 2 2 

c /*sin 2 £cfo sin 3 # . sina; 1, , . , x 

6. I — — - — = - t- + ~ -„ -logfseccc-f- tana;). 

J cos°a; 4 cos 4 a; 8 cos 2 a) 8 

J tan 2 a?-l 4 5 ^ 4 

/~tan(# + a)da; , , , , x 

8. I * — ! — l — = x — tan a log (cot x — tana). 

%J tian a? 

/* c?a? &a; , a , , . . , J 

9 - J ^^T- 6 = ^T^ + ^i^ log(asma: + 6cosa;) - 



31. Trigonometric Formulas of Reduction. 

By means of the following formulae, I sin m a;cos n a;da,* may be 

obtained for all integral values of m and n, by successive re- 
duction. 

■xdx 



I sin TO a;cos n i 

sin m_1 xcos n+:i a; , m — 1 C m <> „ 7 /^ 

= 1 I sm m_2 a;cos n a3c?a;. . (1) 

m + n m + nJ 

J cos n xdx _ cos w+1 a; m—n—2 rcos n xdx /„\ 

sin"* a; (m — l)sin m_1 a; m — 1 J sin" 1-2 a; 

J sin m a;cos w a:da; 

sin m+1 cccos M - 1 a; . n — 1 C • «» n-s ^ /ox 

= H I sm^acos" 2 a;da;. . . (3) 

m -f % m + 9W 

/sin m a?da?_ sin m+1 a? n— m— 2 rsm m xdx ,.^ 

cos n a; (n — l)cos' i-1 a; n— 1 J cos n ~ 2 a; 

/■ _. , sin" 1-1 x cos a; . m — 1 /* . _ « 7 ^n 

sm m £cda; = — ■ 1 I sin m-2 a;cfo. ... (5) 
m m J 



TRIGONOMETRIC INTEGRALS. 225 

/ dx cos a; ,m — 2 C dx ,„. 

sin m # — (ra — l)sin m_1 a; m — 1 J sin m - 2 ic 

/„ 7 sina3Cos n_1 £c , n — 1 f* n _ 2 -, /r7X 

cos n xdx = 1 I cos w 2 xdx (7) 
n n J 

Jdx _ sin a; n — 2 /* da; xgx 

cos n a; (w — l)cos n_1 a; n — lJ cos n ~ 2 # 

32. Derivation of the Formulce in the preceding article. 
To derive (1), we integrate by parts with u = sin" 1-1 **;. 

/■ ,, n j sin m_1 ajcos n+1 aj , m — 1 C • « 2 «+2 7 

n+1 n+lJ 

I sin m-2 a?cos n+2 a;da? = I sin w_2 a;cos n a;da; — J sin w a?cos n a;da;. 

Substituting this in the preceding equation, we have 

(m + n) J sin w a; cos M a; dx 

= — sin m-1 a?cos n+1 a5 + (m — 1) | sin w - 2 a? cos n cc da?, 
which gives (1). 

To derive (2), substitute in (1) 

m — 2 = — m', m = 2 — m', 

afterwards omitting the accents and transposing. 

To derive (3), integrate by parts with u = cos n ~ 1 x, and pro* 
ceed as in the derivation of (1). 

To derive (4), substitute in (3) 

n — 2 = — n', n = 2 — n\ 
afterwards omitting the accents and transposing. 

To derive (5) and (6), make n = in (1) and (2), respectively. 
To derive (7) and (8), make m = in (3) and (4), respectively. 



226 INTEGRAL CALCULUS. 

EXAMPLES. 

i Cc^e^/j^ cos£c/sin 5 a; ,5-o , 5 • \ , 5x 
1. J.smxto = - — [—+-8uto + -Bmx)+- 

J 4 ^sm^ 2sinV 8 5 2 

J 2 cos 2 a \3 cos 4 cc 12 cos 2 oj 8y 



+ — log (sec x + tan a?) . 

4. fcos 8 a;cto=— (Wa+-cos 5 a+^ cos 3 z+^cosaA+^. 
J 8 V 6 24 16 y 128 

K r • 4 2 ^7 cos x /sin 5 a? sin 3 ic sinaA . x 

5. I sm 4 a; cos 2 # ax = H ■• 

J 2 V 3 12 8 y^l6 

6. I — — — = : — cos 4 # -f sin^c cos 2 # -(- - sm^ 

J siirx sina;\ 2 J 2 

coscc /0 2 \ 3cc 



(3 — cos 2 #) 

2sm^ v J 2 



, r cto = if 1 5 5 8in \ 

*/ sin 4 a?cos 3 a; cos 2 a; \3 sin 3 x 3 sin a? 2 / 



5 
+ o l°g ( sec # + tan %) ■ 



33 



L Required J 

*/ a -f 6 si 



since 



a + 6sinaj= a( cos 2 - + sin 2 - j+26sin-cos — 

V Zi ZJ Z Li 

sec 2 -dec a sec 2 - da? 

dx r 2 r 2 



/ax r L _ r* 

a+bsmx-J a + 2bt ^ + at a n% = J jfatanf+ftY+tf-* 

= 2 f I , where z = atan^ + 6. 
J z 2 -f- a 2 — 6 2 2 



TBIGONOMETEIC INTEGRALS. 227 



If a>h, numerically, 

dx 2 



J a + 



tan~ 



b sin x Va 2 -6 2 Va 2 -6 2 

x 
a tan- -\-b 
2 2 

— tan -1 - 



Va 2 -6 2 Va 2 -6 2 

If a<b, numerically, 



dx C dz 1 i z—Vtf—a 2 



c dx = 2 r — * — = l log 

J a +& since «/ z 2 — (5 2 — a 2 ) V& 2 — a 2 



z 2 -(& 2 -a 2 ) V6 2 -a 2 2 + V6 2 -a 2 



a tan - + 6 — V& 2 — a 2 
log 



V& 2 - a 2 atan ^ + 6 + V6 2 - a 2 



34. Required I - 
J a 



H-frcosa 



a + 6 cos cc = a [ cos 2 - + sin 2 - ] -f 6 [cos 2 - — sin 2 - ] 
V 2 2) \ 2 2) 



= (a + 6)cos 2 - + (a - 6)sin 2 -- 



/dx r 

a -f- 6 cos a? ~~ J 



sec 2 -cfo» 



a-\-b + {a — 6)tan 2 - 
Z 



x 
If we put tan- = z, 

A 

/ dx _ r dz 2 r dz 

a + 6cosa;~ J a + b-\-(a — b)z 2 ~~ a — bJ 2 . a + b 

a — b 
If a>b, numerically, 



r dx = _2 J«-j tan 

«/a-J-&cosa; a — 6*a + Z> 



&fa«-i*"va-& 



2 



tan -1 f -v /- 1 an - V 

\A/a + 6 2) 



228 INTEGRAL CALCULUS. 

If a <b, numerically, 

/ dx _2__ r dz 

a + bcosx~ 6 — a J 2 b + a 



b — a 



1 1( zVft-a-V& + a 
V& 2 — a 2 zV& — a + V&-fa 



V& — a tan- +V6 + 



:log- 



V& 2 -a 2 V6-atan|-V6T^ 

c . r ^ 

35. Required I e^sinn&da?, and I e^coswajaa;. 

Integrating by parts, with u = e™, 

/«„ • 7 e ax cosn« , a /• „, 7 ... 

e ax sm wa?a# = h- I e ax cosw#a#. . (1) 
w nj v 

Integrating the same, with u = sinwa;, 

J»„ . 7 e ax sinw# n r nT . /rkN 

e az smna;c?ic = I e ax cosmca:r. . . (2) 

Eliminating from (1) and (2) I e°* cos nxdx, we have 
(a 2 + n 2 ) j e ax sin nx dx = e ax (a sin nx — n cos na) *, 

v C ax • j e™ ( a sin nx — n cos nx) 

hence I e^ sin nee asc = — * '- • 

J a 2 + n 2 

Substituting this in (1) and transposing, gives 

a C ax j e ax {ans\nnx-\-a 2 cosnx) 

- I e ax cos nxdx — — * — -; 

nJ (ar + n^n 

i, C n.x j e ax (nsmnx-\- acosnx) 

hence I e ax cos nx dx = — ^ — — '— 

J a 2 -h n 2 



TRIGONOMETRIC INTEGRALS. 229 



EXAMPLES. 



tan 2 



1. f *» =ho g 2 . 
J 4-5sin* 3 o^*^ 

2 

2 r dx _ 1 

J 5 + 4 sin 2 a; ~" 3 



dx 1, _i5tanic + 4 
= -tan l : — • 



tan- + 2 

3. c d * =ho g 2 . 

J3 + 5cosz 4 tan ^_ 2 

2 

4# r d» = ltan- 1 / r 2tan?Y 

J 5-3cosz 2 V V 

5. f da = ltan- 1 (3tanaj) 

J 5-4cos2a 3 v ' 

J_ /m ~ f . x x\ 

e 2 cos-dx = e 2 ( sin- -f cos- J« 

» /^ sin a; c?a; _ _ sin a? 4- cos x 
J ? ~" 2e s 

T x • 2 t eV-i 2 sin 2 a 4- cos 2x y 

8. I e x sin 2 ajc?x = -( 1 £- 

n r x ■ o • j e* / . . 3 sin 3 cc + cos 3 cc N 

9. I e x sm2a?sin£ca£ = — [smaj + cos# - 1 

10. I e ax (sinaic + cosaa;)cZx 



c^smaa; 



a 

See 



Je 
e 3 * (sin 2 a? — cos 2 a) cte = — (sin 2 <c — 5 cos 2 a). 



CHAPTER VI. 

INTEGRALS FOR REFERENCE. 

36. We give for reference a list of some of the integrals of 
the preceding chapters. 



x n dx = ^—- 
re + 1 

2. r**=io g *. 

J x 

3. r *L_ = itMi-?. 

»/ ar -f- or a a 



/; 



efc» 1 ■> x — a 



x 2 — a 2 2 a x + a 
Exponential Integrals. 

5. fa*cto = — • 
J log a 

6. Ce*dx = e x . 

Trigonometric Integrals. 

7. I sin x dx = — cos x. 

8. I cos x dx = sin a?. 

9. I tan x dx = log sec x, 
10. j cot xdx = log sin cc. 



INTEGRALS FOR REFERENCE. 231 

11. I secxdx = log(seca?-ftana;) 

= logtang + |). 

12. I cosec#cZa; = log(coseca; — cota;) 

= log tan ^- 

13. j sec 2 x dx = tana;. 

14. I cosec 2 xdx = — cot a;. 

15. j secxt&nxdx = sec a;. 

16. j cosec£ccotcccfcc = — coseca;. 

17. lsm 2 xdx = sin2a?. 

J 2 4 

18. ( cos 2 icdic = - + -sin2o;. 
J 2 4 



Integrals containing Va 2 — a 2 . 

19. I - = sin 1 — 

*^ Va 2 — ic 2 a 

20. I — :IZ== = — V a 2 — x 2 H — sm 1 - 
J Va 2 - a 2 2 2 « 



21 



. r_^ == i log . 

*^ x Va 2 — # 2 a i 



Va 2 — a; 2 a a + Va 2 — x 2 

7; 



22 C dx - *^°^ ~ x ' 2 

J a; 2 Vet 2 - a 2 ~~ ^ x 



23. f dx = - Va2 ~ a;2 4--J-lo 

J ^Va 2 — ic 2 2a 2 a; 2 2<x 3 



2aV 2a 3 ° ft + v '^^ 



232 INTEGRAL CALCULUS. 

24. Cy/a*-a?dx = *-Vtf—tf + ^sin- 1 -. 
J 2 2 a 

. fxWa 2 -a?dx = -(2 ar 2 - a 2 ) Va'-x 2 + - sirr 1 -. 
*/ 8 8 a 



25 



26. f 



(a 2 - x 2 f a 2 Va 2 - x 2 

27. C(a 2 -x 2 )%dx = x : (5a 2 - 2x>) V^^ +— sin" 1 -. 
«/ 8 8 a 

Integrals containing Var + a 2 . 

28. f da; = log(s+ Vg + a»). 

•^ Va^ + a 2 

29. f x ' dx =gVa?T^-^log(a?+V^T^). 

^V« 2 + a 2 2 2 



30 



/dx 1-, # 

— = -log 

a; Vic 2 + a 2 a a + V # 2 + a 2 



31, 



/: 



dx Vic 2 + a 2 



x^^/x 2 + a 2 



a + vaT+a 2 



J a? Va; 2 + a 2 ~ 2a ^ 2a 3 ° g 

33. fVx T +~a 1 dx = | Va?T^~ 2 + % log (a? + Var 2 + a 2 ). 

34. fa; 2 Va?+a 1 da;=f(2a^+a 2 )V^^ 

•/ 8 8 

35. f_gg_ = g . 

J (ar^ + a 2 )* aV^ + a 2 

36. f(af~+a 2 ) f (fa==f(2^+5a 2 )ya?^^ 

•/ 8 o 



INTEGRALS FOR REFERENCE. 233 



Integrals containing V x 2 — a 2 . 

37. f dx =log(^+V^=^). 
•^ Va; 2 — a 2 

38 r x 2 dx = gv^r^ + ^iog(a, + v^z^). 

*^ Va; 2 — <x 2 ^ ^ 



dec _ 1 

a; Var 2 — a 2 a a 



39. f— 



40. 



/: 



c?a; Va; 2 — a 2 



a; 2 Var 2 — a 2 a * 



-.. /* da; -yx 2 — a 2 , 1 ,a? 

41. = 022 + ^~3 SeC - 

J tf-y/x 2 — a 2 2 ax 2 or a 

42. f V^^a 1 da; = | Va^~a~ 2 - ^ log (a; + Va; 2 - a 2 ) . 

43. fa; 2 Va^^ 2 da;=|(2a; 2 -a 2 )V^^ 2 -^og(a;4-Va^^o?) < , 

»/ 8 8 

44 r__o^_ = _ x 

' J (a? _ «)t a 2 Var 2 - a 2 

45. C(x 2 -a 2 ^dx=-(2x 2 -5a 2 )^/x T ^a~ 2 +— log^+Vtf 211 ^) 
«/ 8 8 



Integrals containing ■\/2ax—x 2 . 



CIX _i X 



46. 1 — — = vers -1 

J . 



■\/2ax — 



2 a 



47> fjs<fo__ = _V2aa;-a^ + avers- 1 -- 



V 2 aa; — a; 2 



48 



■/: 



dx V2. 



a; V2 aa; — a; 2 ax 

49. I V2 aa; — a; 2 da; = ~ V2 ax — x 2 -\ — vers -1 -' 
J 2 2 a 



234 INTEGRAL CALCULUS. 

ca C /o 9 j 3a 2 -\-ax—2x 2 /„ s . aS -i x 

50. I xW2ax— x 2 dx = ±— V2aaj— a^+ — vers '— 

J 6 2 a 



K1 r^/2ax — x 2 dx /77~ — ~2 > -n 

51. I = V2 ax — ar-f a vers J 



^ ; V2 aa; - a; 2 <fa = (2 ace - x 2 )* 
x 3 Sax 3 



53 ( * dx _ _ eg — a 

^ (2a«-f)^ aWZax — x 2 

54. I = — 

^ (2aa> — a 2 )* a-y/2ax — tf 

Integrals containing ± a# 2 -}-&# + c. 

55. f ^ = 

J aar 2 + 6a; + 



, 1 tan-' 2ax + b 

G V4 ac — b 2 V4 ac — b 



56 . or „ i ^2^ + 6 - yy-^ 

V& 2 — 4 ac 2ax + b-\r V& 2 — 4 ac 

57. f da? — = -Llog (2 aa; + 6 + 2 VaVaar* + bx + c). 

•^ Vaar 2 + 6a; + c Va 

58. ( ^/ax 2 -\-bx-\-cdx = ■ Vaa; 2 + bx 4- c 

J 4a 



& 2 — 4aci 



■j" 



__u — l^io g (2ax+6+2VaVaa; 2 +da;-|-c). 

8a^ 

da; 1 . _ x 2ax — b 



59. — = -i-sin 



V—axP + bx + c Va V& 2 + 4ac 

60. J V— ax* + bx + cdx 

2ax—b / or? — r~ . ^-Mac . _i 2 ax—b 

= — V — aa; 2 +&a;-fc-f- — sm 

4 a 8a | V6 2 +4ac 



61. CJ±±*dx 



INTEGRALS FOR REFERENCE. 235 

Other Integrals. 



= -y/(a+x) (6+ a;) + (a — 6)log( Va+x+->/b+x). 
62. CJ^~^dx = V(a-x)(b+x) + (a + Msm-^/^iA 



CHAPTER VII. 



INTEGRATION AS A SUMMATION. 
INTEGRALS. 



DEFINITE 



37. The process of integration may be regarded as the 
summation of an infinite series of infinitely small terms. 
As an illustration, consider the following problem. 

38. To find the area 
PABQ included between 
a given curve OS, the 
axis of X, and the ordi- 
nates AP and BQ. 

Let y = x* be the equa- 
tion of the given curve. 

Let OA = a, OB = b. 

Suppose AB divided 
into n equal parts (in the 
figure, 7i — 6), and let Ax 




Ai A 2 A 3 A 4 A 5 B 



denote one of the equal parts, as AA 1} A X A 2 , ••*. 
Then AB — b — a = nAx. 

At A l} A 2 , •••, draw the ordinates A^P^ A 2 P 2 , 
plete the rectangles PA X , P X A 2 , • ••. 

Prom the equation of the curve, y = x*, 



>, and com- 



QB = b\ 



PA = a*, P X A Y = (a + Ax)*, P 2 A 2 = (a + 2 Ax)*, 
Area of rectangle PA 1 = PA x AA X = a* Ax. 
Area of rectangle P^A 2 = P 1 A 1 x A X A 2 = (a + A#)^A#. 
Area of rectangle P 2 A 3 = P 2 A 2 x A 2 A 3 — (a + 2 A#) * A#. 



236 



INTEGRATION AS A SUMMATION. 237 

The sum of all the n rectangles is 

c^Ax + (a + Aa>)* Ax + (a + 2 Ax)s Ax H (- (6 — Ax)^ Ax, 

which may be represented by A^^Aa;, 

where x^Ax represents each term of the series, x taking in suc- 
cession the values a, a + Ax, a -f 2 Ax, •••, 6 — Ax. 

It is evident that the area PABQ is the limit of the sum of 
the rectangles, as n increases, and Ax decreases. 

When Ax in the preceding series is changed into the infini- 
tesimal dx, the symbol ^ is replaced by J , an abbreviation 
of the word " sum." 

Thus 

Xx^dx = b?dx + (a + dx)^dx + (<x + 2 dxftdx -\ \- (p — dx)^dx 

= area PABQ (1) 

The expression I x 2 dx, as defined by (1), denotes the sum of an 

infinite number of terms, each of which is represented by x^x, 
x taking in succession the values a, a -f- dx, a + 2dx, •••, b — dx. 
Or the definition may be more precisely expressed by 

•?dx = Limit of ^^x^Ax, as Ax approaches zero. 

It is to be noticed that a new meaning is thus given to the 

symbol | , which has been defined heretofore as the inverse of 

differentiation. It will now be shown that the two definitions 
are consistent. 

39. Value of I x 2 dx. To find the area PABQ we must find 

the sum of the series (1), Art. 38, that is, the value of I x 2 dx. 

Now x s dx = dl-x s y 

But the differential of a function of x is the increment of 
that function when x receives the increment dx. 



x 



238 INTEGRAL CALCULUS. 

That is, d (| x?) = ^(x + dxf - - x\ 

\y Jo 3 

1 2 3 9 3 

Hence, xHlx = - (x + dx) * x\ 

Substituting for x, a, a -\- dx, a + 2 dx, •••, b — dx, 

1 2 3. 2 3 

we have a?dx = -(a-\-dx)* a*, 

o o 

(a + dxfdx = -(a + 2 dx)% - - (a + dx)% 
o o 

(a + 2 dxfdx = | (a + 3 da?)* - | (a + 2 da?)*, 



(b - dxfdx = § $ - 1 (6 - dz)l 
Adding and cancelling terms in second member, we have 

afdx -f (a + da^dsc + (a + 2 dx)^ H 1- (b — dx)^dx = ? $— - a\ 

3 o 

That is, as x varies from a to b, the sum of the successive 

2 s 
increments of the function - x' 1 is equal to its entire increment. 

o 

Thus Cxhx = 1 6* - ? J = area P^JBQ. 

c/a 3 3 

We have thus shown that the sum of the infinite series repre- 
x 2 dx is found by substituting for x, b and a in 

a ! 

- x % , the integral of a^dx, and subtracting the latter result from 
o 

the former. 

The expression f x*dx is called a definite integral, and the 

process of evaluating it is called integrating between limits, the 
initial value a of the variable being the inferior limit, and the 
final value b the superior limit. 

2 3 

In contradistinction -x 1 is called the indefinite integral of 
x^dx. 



INTEGRATION AS A SUMMATION. 239 

40. The relation of the terms of the series I x*dx to the 

2 s ^ a 

integral - x 2 may be made clearer to the student by consider- 
ing the following series of numbers : 

1 



4 

9 

16 

25 

36 



3 
5 

7 

9 

11 



The numbers in the second column are the differences be- 
tween consecutive numbers in the first, and it is evident that 
the sum of the second column of numbers is the difference 
between the first and last, in the first column. That is, 

3 + 5 + 7 + 9 + 11 = 36-1. 

X5 1 
x 2 dx may be similarly arranged, as follows : 



2 3 

— a^ 

3 ' 






a 2 dx } 


|(o + *»)*. 






(a + dxydx, 


|<« + 2<to)», 






(a + 2 dx)^dx, 


|(a + 3<te)», 




|(&-d#, 




i* 


(b — dx)?dx. 



240 INTEGRAL CALCULUS. 

1 2 3 

Since x 2 dx is the differential of - x J , the terms in the second 

o 

column are the infinitesimal differences between the consecu- 
tive terms in the first, and therefore 

aklx + (a + dx)*dx + (a -f 2 dx) i -\ \-(b— dxfax = -b^ — -a?; 

that is, | x 2 dx = -b 2 a?. 

X 3 3 

41. General Definition of a Definite Integral. 

In general, if <f> (x) denote any given function of x, which is 
finite and continuous from x = a to x = b, I <£ (x)dx is the 

%J a 

definite integral representing the sum of an infinite series of 
terms, obtained from <f>(x)dx, by supposing x to vary from 
a to b. 

If I cj> (x)dx = if/(x) f the indefinite integral, 

then I <£ (x)dx = ip(b)—ij/ (a). 

This may be illustrated by an area as in Art. 38, by suppos- 
ing y = <fi(x) to be the equation of the curve OS, and the proof 
of Art. 39 may be similarly modified, by substituting <£ (x) for 

i 2 i 

x* f and ij/(x) for -x 2 . 
o 

42. We add in this article the proof of the relation between 
the definite and indefinite integrals, expressed in the form of 
limits instead of infinitesimals as in Art. 39. 

We shall use the expression " Limit Ax=0 " to denote the words 
" The limit, as Ax approaches zero, of." 

Given <j>( x ) = — ( x )> an ^ 

ax 

2* ( x ) & x = <l> (a) Ax + <f> (a + Ax) Ax + <£ (a + 2 Ax) Ax + ••• 

+ <f> (b — Ax) Ax 3 



DEFINITE INTEGBALS. 241 

the function <£(x) being finite :.^d continuous from x = a to 
x = b ) to prove that 

IAmit± x= $^J>(x)&x== <K^)~~ ^( a )- 
From the definition of —\p{x), in Art. 10, Dif. Cal., 

CtX 

»(») = 4-K») = LimiW =0 ^ (x + Ag) ~ »<*> . 

dec Ax 

Hence *(« + **)-*(*) = ^) + € , 

where e is a quantity that vanishes with Ax. Hence, 
i^(x + Ax) — tf/(x) = <£(x)Ax-feAx. 
Substituting in this equation for x, 

a, a + Ax, a -f- 2 Ax, • • • 6 — Ax, 
we have 

\f/(a + Ax) — ip(a)= </>(a)Ax-t-e 1 Ax, 

^(a + 2Ax) — i//(a -f- Ax) = <j>(a -f Ax) Ax -f- e 2 Ax, 

^(a + 3 Ax) — if/ (a + 2 Ax) = <f>(a + 2 Ax) Ax -f e 3 Ax, 

^(5) _ ^(J _ Ax) =4>(b — Ax) Ax + e n Ax. 

Adding and cancelling terms in first member, we find 

^(6)-«A(a) = 5)>(x)Ax + ^/Ax (1) 

Now if e k is the greatest of the quantities c^ e 2 > •■• c n> it fol- 
lows that 

^/Ax<e fc ]^*Ax; 
that is, V a eAx< £^(6 — a), 



242 INTEGRAL CALCULUS. 

Hence ^V eAx vanishes with. e k9 that is, with Ax. 
Taking the limit of (1), we have 

i[/(b) — ijs (a) = Iuim.it & x = o^ a 4> (%)&%== J <f>(x)dx. 

43. It is to be noticed that the arbitrary constant c, in the 
indefinite integral, disappears from the definite integral. 

Thus, if in evaluating j x*dx, we call the indefinite integral 

SC 4 

— h c, we have 
4 

C h a?dx = ^ + c -(°t + cV ? - ^ as before, 
•/a 4 \4 / 4 4 

Or if in evaluating J <j>(x)dx, we call the indefinite inte- 

gral \p(x)-\-c, we have 

j*4>(x)dx = $(b) + c - |>(a) + c] = ^(6) - ^r(a), 

as before. 

EXAMPLES. 



Evaluate the following definite integrals : 

Ji 3|i 3 3 

2. f— = log a; '= loge - logl = 1. 
J\ x 1 

IT IT 

J^2 12 

sin xdx — — cos a; = — ( — 1) = 1. 
o [a 

I (b 2 x — x s )dx = — • 



4 



5. r^=i. 



DEFINITE INTEGRALS. 243 



« f s xdx = log2 
* J2 1 + flj 2 2 

»/o or -|- 4a 2 

8. f%ec 4 0d0 = i. 
Jo 3 



9. I sclogsccfcc 



e 2 + l 
4 

1+V2 > 



10. f^ = logfl±V2 > ) 
•/tt cos a; \ -y/3 / 

6 



jc 2 — 2cccosa + 1 2sina 

12. f *? = » 

Jo (a 2 + a 2 )(& 2 + a 2 ) 2a6(a + &) 

Jr 00 1 

e~ na: sinft#cfo = — • 
o 2n 

IT 

14. f' ^ = g__ . 
Jo 2 + cosa; 3^/3 

Derive the following by (5) and (7), Art. 31: 

15. If w is even, 

Csm n xdx = Ccos n xdx = 1-3-5 — fo-1) 
Jo Jo 2-4.6.. -n 

16. If n is odd, 



J" 2 r 2 2'4:' 6 •••( 
sm tt ajc?a?= I cos M #cfcc = — ^ 
o Jo 3-5.7. 



2.4-6. >. (n-l) 



244 INTEGRAL CALCULUS. 

43|-. Change of Limits. When a new variable is used in 
obtaining the indefinite integral, we may avoid the restoration 
of the original variable, by changing the limits to correspond 
with the new variable. 

For example, to evaluate 

J" 4 dx /- 
-, assume V a; = z. 
1 4-a/o! 



Then we have 



r ° 1+Vx 

dx 2zdz 



1+Va; ! + z 
Now when x = 4, z = 2 ; and when x = 0, z = 0. 

Hence C dx = f 2 ?A^ = 2 [z-log(l + z)] 2 
= 4-21og3. 

EXAMPLES. 

1. f'(*- *)**» = 6. Assume „ = L 

Ji a: 4 z 



2. I j L = 8 H — tt- Assume # — 2 = z 3 

*/3 /„. 9\3 I Q 2 



i^Lf ±cto = 4-«-. Assume e*-l = z 2 . 

o e* + 3 

4. p ^ — =-J— log (sec a + tana). 
Jo -^/ e 2x + tan 2 a tan a 

Assume e~ x + tan 2 a = z 2 . 

5 p(sinl+cp^)^ = log3. Agsume sin ^_ cos ^ % 
Jo 3 + sm26> 4 

6 r 2 + ^ (« 8 + l) ^_ = l0 g 3> Assume x-- = z. 

J 1 Wa3 4 + 7X 2 + 1 * 



CHAPTER VIII. 



APPLICATION OF INTEGRATION TO PLANE CURVES. 
APPLICATION TO CERTAIN VOLUMES. 

44. Areas of Curves. Rectangular Co-ordinates. The sim- 
plest application of integration to curves is in determining the 
areas defined by them. We have already used this problem in 
Arts. 38, 39, as an illustration of a definite integral. We shall 
now consider it in a more general form, and derive the expres- 
sion for the area included by any curve, in rectangular co- 
ordinates. 

45. To find the area between a given curve PQ, the axis of 
X, and two given ordinates, AP and BQ. 

Let OA = a, OB = b. 

Let x and y be the co- 
ordinates of any point 
P 2 of the curve ; then 

x + Ax, y + Ay, 

will be the co-ordinates 

ofP 3 - 

The area of the rec- 
tangle P 2 A 2 A 3 is 

P 2 A 2 X A 2 A Z = yAx. 

The sum of all the rectangles PAA Y , P X A X A 2 , P 2 A 2 A 3 , ••-, 

may be represented by ]v yAx. 

The required area PQBA is the limit of the sum of the rec- 
tangles, as Ax is indefinitely diminished. That is 







= I ydx. 



246 INTEGRAL CALCULUS. 

We may also regard the required area as generated, by the 
ordinate AP moving from left to right, and varying in length 
according to the equation of the given curve. Regarding y as 
constant while moving the distance dx, it generates the rec- 
tangle ydx. Then the general formula for the required area is 



A = I ydx, as before ; 



the inferior limit a — OA, denoting the initial position of the 
moving ordinate, and the superior limit b = OB, its final posi- 
tion. 

Similarly the area between the given curve, the axis of Y, 
and two given abscissas, is 

A=jxdy, 
the limits of integration being the limiting values of y. 

EXAMPLES. 

1. Find the area between the parabola y 2 = 4: ax and the axis 

of X, from the origin to the ordinate at the point (li, k). 

Here A= \ ydx= \ 2a I x I dx = — - — = — - — 
c/o Jo 3 o 3 

Since k 2 = 4,ah, k = 2ah%. 

2 
.*. A = -hk, two-thirds the circumscribed rectangle. 
o 

2. Find the entire area of the ellipse — \- % = 1. Arts. nab. 

a 2 b 2 

3. Show that the area of a sector of the equilateral hyper- 

bola x 2 — y 2 = a 2 , included between the axis of X and a 
diameter through the point (x, y) of the curve, is 

a 2 -, x -f y 
log -^JL. 

2 a 

8 a 3 

4. Find the entire area between the witch y= — -, 

and the axis of X. 

Ans. 47ra 2 . 



ABE AS OF CURVES. 



247 



5. Find the area intercepted between the co-ordinate axes by 



the parabola x* + y 2 = a 2 . 



Ans. 



6. Find the entire area within the curve (-) 4- f ^ V= 1. 

WW 

Ans. -wabo 
4 

7. Find the entire area within the hypocycloid x 3 + y 3 = a 3 . 

3ira 2 



Ans. 



8. Find the entire area between the cissoid y 2 = 
and the line x = 2 a, its asymptote. 



X s 



a\ 



2a — x 
Ans. 37ra 2 . 

The area between two curves is the sum, or the difference, 
of the areas between the curves and one of the co-ordi- 
nate axes, the limits being determined by the points of 
intersection. 

9. Find the area included between the parabola x 2 = 4 ay, 

8 a 3 / 4\ 

and the witch y = — Ans. [2ir ] 

9 ^ + 4a 2 V 3 / 



. 46. Areas of Curves. 
Polar Co-ordinates. To find 
the area POQ, included be- 
tiveen a given curve PQ, and 
two given radii vectores, OP 
and OQ. Let 

POX=a, QOX = (3. 

Let r and be the co-or- 
dinates of any point P 2 of 
the curve, then 

r + A?-, 6 + A<9, 

will be the co-ordinates 
ofP 3 . 




248 INTEGRAL CALCULUS. 

The area of the circular sector P 2 OR 2 is 

1 0P 2 X P 2 R 2 = |r • rA0 = ir*A0. 

The sum of the sectors P0i2, PxOR^ P 2 OE 2 , • ••, may be rep- 
resented by p -• 

The required area POQ is the limit of the sum of the sectors, 
as A0 approaches zero. That is, 



=D> 



47. We may also regard the area POQ as generated, by the 
radius vector revolving from OP to OQ, and varying in length 
according to the equation of the given curve. 

Regarding r as constant while describing the angle dO, it 

generates the sector whose area is -i^dO. 



1 C^ 
Hence A = - I i^dO, as before ; 

2*J o. 



the inferior limit a denoting the initial, and the superior limit 
(3, the final position, of the moving radius vector. 

EXAMPLES. 

1 . Find the area described by the radius vector in one entire 
revolution of the spiral of Archimedes r = aO. 

2 - 47r 3 a 2 



1 /*2tt 1 /»2tt n 

Here A = ± I r 2 dO = ±l a 2 2 d0=- 

2 Jo 2 Jo 2 



23 



2. Find the area described by the radius vector in the loga- 
rithmic spiral r = e ae , from = to = ~- 

Ans. — (e na — l). 
4a 



«*$ 



LENGTHS OF CURVES. 



3. Find the entire area of the circle r = asin0. 



249 



Ans. **. 



4. Find the area of one loop of the curve r = a sin 20. 



Ans. ** 
8 



5. Find the entire area of the cardioid r=a(l — cos#). 

Ans. ——-, or six times the area of the generating circle. 

6. Find the area described by the radius vector in the parab- 



ola r=asec 2 -, from = to = -» 

2' 2 



Ans. 



4 a' 



7. Find the area below OX within the curve r = a sin 3 — 

3 






Ans. (10tt + 27V3) 



G4 



48. Lengths of Curves. Rectangular Co-ordinates. To find 
the length of the arc PQ between two given points P and Q. 

Let OA = a, OB = b. Y Q 

Denoting the required length 
of arc by s, we have 

'dy 



ds = 



i+ 



(1) Art. 98, Dif. Cal. 



therefore 
s = 



rHDi 



dx; 




the limits of integration being the limiting values of x. 
Or we may evidently use the formula 



/[ 



■'I 



dy, 



the limits being the limiting values of y 



: ^ 



250 INTEGBAL CALCULUS. 



EXAMPLES. 



1. Find the length of the arc of the parabola y 2 = 4 ax, from 
the vertex to the extremity of the latus rectum. 

Here .& = ( t 

dx Jr 



therefore , = J Yl + g**, = J] Y^T*» 

This may be integrated by 9, p. 213, making 6 = 0. 

J [ ^lZ_^ ydx= Vaa>4-a^ + alog(Va + a;4--Va;). 

jf Y a + x \*dx = a[ V2 + log (1 + V2) ] = 2.29558 a. 



2. Find the length of the arc of the semi-cubical parabola 
ay 2 == X s , from the origin to x = 5 a. . 335 ct 

/lS ' 27 ' 



3. Find the length of the arc of the curve 9 ay 2 = x(x — 3a) 2 
from x = to x = 3 a. Arts. 2 a V3 



4. Find the length of the arc of the catenary y = - (e a + e a ), 

from x = to the point (a;, y) . 

X X 

2 V 

5. Find the entire length of the arc of the hypocycloid 

2 2. 'A 

x 3 4- y 3 == a 3 . Ans. 6 a. 



49. Lengths of Curves. Polar Co-ordinates. To find the 
length of the arc PQ between two given points P and Q. 



Let POX=a, QOX=j3. 



LENGTHS OF CUBVES 
We have 



dOJ 



(10 



therefore 
s 



-r 



(3) Art. 98i, Dif. Cal 



idt 
i + 
1 d$ 



the limits being the limiting values 
of 0. 



Or we have ds = 1 -f- r 




<W, (1) 



o/W 



therefore 






(2) Art. 984, Dif. Cal. 



(2) 



the limits being the limiting values of r. That is, OP = a, 
OQ = b. 

EXAMPLES. 

1. Find the length of the arc of the spiral of Archimedes 
r = aO, from the pole to the end of the first revolution. 
dr 



Here 



dO 



= a. 



d$ 



s = C \a?6 2 + a 2 fd0 = a f *(1 :'■+ 2 )^ 

= a [^? + |i og(e+ VTT#)p 

= ar7rVl + 47r 2 + ilog(27r+Vl+47r 2 )l. 

2. Find the entire length of the cardioid r = a (1 — cos #). 

3. Find the length of the logarithmic spiral r = e ae , from the 

pole to the point (r, 0). Use formula (2). 



Ans. -Vtt 2 + 1. 
a 



252 



INTEGRAL CALCULUS. 



4. Find the entire length of the curve r = a sin 3 



Ans. 



Sira 



5. The equation of the epicycloid, the radius of the fixed 
circle being a, and that of the rolling circle -, is 

s i n 2 q = 4(V-cr) Find the j t k of one loop. 
27 oV 
From the above equation 



d0 2vV - a 2 



dr 



■V4a 2 



; then use Formula (2). Ans. 6 a. 



50. Surfaces of Revolution. Volume. To find the volume 
generated, by revolving about OX the plane area APQB. 

Let OA = a, OB = b. 







c 


) ^^*^* Let - a; and y be the 






pj^\ 


co-ordinates of any point 




p 


V^a\ 


P 2 of the given curve. 




K^ 


i I 

I I 


It is evident that the 


p 


i 
i 


I 
i 
i 

! 

! 1 

l 

| 1 

i 


rectangle P 2 A 2 A 3 will 
generate a right cylin- 
der, whose volume is 
iry-\x. 






i 

i I 


The sum of all these 
cylinders may be repre- 


J 


^ A r I 


^2 A 3 A 4 [ 


*■ sented by ttA y 2 Ax. 



I 



The required volume is the limit of the sum of the cylinders, 
as Aic approaches zero. That is, 



V=irCy 2 dx. 



Or we may regard the required volume as generated by the 
area of a circle, which moves with its plane always perpendicu- 
lar to the axis of X, its centre moving along this axis, and its 
radius being the ordinate of the given curve. 



SUBFACES OF REVOLUTION. 253 

Since y is the radius of this moving circle, its area is try 2 , 
and regarding y as constant while it moves over the distance 
dx, we have for the volume of an elementary cylinder, 

dV '= ivy 2 dx. 



Hence 



= Trfy 2 dx, ....... (1) 



the limits being the limiting values of x. 
Similarly, if Y is the axis of revolution, 

V= 7r I x 2 dy, 

the limits being the limiting values of y. 

51. Surfaces of Revolution. Area. To find the area of the 
surface generated, by revolving about OX the arc PQ. 

In the figure of Art. 50, let P 2 P 3 be an element of the given 
curve. 

This will generate the convex surface of the frustum of a 
right cone. Hence we have by geometry, for an element of 
the required surface, 

dS = 2 ^ A * + P ^ W 3 = ir(y + y')ds, 

where y=P 2 A 2 , and y f = P S A 3 . 

But since the limit of y 1 is y, we have 
dS = 2iryds; 

hence S = 2*1 yds. 

By (1) Art. 98, Dif. Cal., 



Similarly if Y is the axis of revolution, 
S=*2tc i xds. 



dx (1) 



254 



INTEGRAL CALCULUS. 



EXAMPLES. 



lo Find the volume and surface of the prolate spheroid 

obtained, by revolving about X the ellipse — f- ^- = 1. 

a 1 b- 



From (1) Art. 50, we have 



-V=7r I y 2 dx == 7r ( — (a 2 — x 1 ) dx 
2 Jo u Jo a? K } 



27rab 2 



: V = 



kirab 2 



From (1) Art. 51, 



2/U + 



dx 



1 + 



b 2 x 2 



Jo a a 2 (a- — ar) 

= 2tt- 9 f\a* - (a 2 - b 2 )x 2 fdx 
a- Jo 



2 dx 



= Mb + 



A 



■sin 



_! V« 2 — 6 : 



V<r 



a 2 j6 

cos - 



V Va 2 - 6 

2. Find the volume and surface generated, by revolving about 

X the parabola y 2 = 4 ax, from the origin to x = a. 

Ans. 27ra 3 and 8 (V 8 - 1 ) 7ra 2 < 
3 

3. Find the volume and convex surface of the right cone 

generated, by revolving about X the line joining the 
origin and the point (a, 6). ^ «aP and wbV ^f^. 

o 

4. Find the entire volume and surface generated, by revolving 

2 2 2 

about X the hypocycloid x* + y 3 = a 3 . 

^Itis. 2±£« and M?EL 



105 



OTHER VOLUMES. 255 



5. Find the entire volume generated by revolving the witch 

So? 
x 2 + ±a 2 



y = about X, its asymptote. Ans. 47r 2 a 3 . 



6. Find the volume generated by revolving about X, the part 

of the parabola x 2 +y* = a?, intercepted by the co-ordi- 
nate axes. A 7ra s 

Ans. -■ 

7. Find the volume and surface of the torus generated by 

revolving about X, the circle x 1 + (y — b) 2 = a 2 . 

Ans. 2ir 2 a 2 b and 47r 2 a&. 

8. Find the volume and surface generated by revolving 

X X 

about T, the catenary y = - (e a +e a ), from x=0 to x=a. 

z 



ttOC 



Ans. — (e + 5e- x - 4) and 27ra 2 (l - e" 1 ). 
2i 



52. O/fter Formes. The method of finding the volume of 
a solid of revolution in Art. 50, by considering it generated 
by a moving circle of varying radius, may be extended to any 
solid, where the area of a section can be expressed as a 
function of its perpendicular distance from a fixed point. 




If we denote this distance by x, and the area of the section 
by X, we have for the volume, 



■/ 



Xdx. (1) 



256 



INTEGRAL CALCULUS. 



EXAMPLES. 



Find the volume of a pyramid or cone having any base 
whatever. 

Let A be the area of the base, and h the altitude. 

Let x denote the perpendicular distance from the vertex, 
of a section parallel to the base. Calling the area of 
this section X, as in (1), we have by solid geometry, 

A h 2 ' h 2 ' 

Substituting in (1), 

v= Ar> Mx AV = Ak m 

h 2 Jo h 2 3 3 

Find the volume of a right conoid with circular base, the 
radius of base being a, and altitude h. 

OA = BC = 2a, BO=CA = h. 

The section RTQ, perpendicular to 
OA, is an isosceles triangle. 

Let x = OP ; then 




X = area RTQ = PTx PQ=hW2ax-x 2 . 
Substituting in (1), we have 



V=h 



X2a 
■\'2ax- 



xtdx 



ira 2 h 

2 



This is one-half the cylinder of the same base and 
altitude. 

3. A rectangle moves from a fixed point, one side varying 
as the distance from this point, and the other as the 
square of this distance. At the distance of 2 feet, 
the rectangle becomes a square of 3 feet. What is 
the volume then generated ? Ans. 4£ cubic feet. 



OTHER VOLUMES. 257 



On the double ordinates of the ellipse — |- *L = i and in 

a 2 6 2 

planes perpendicular to that of the ellipse, isosceles 

triangles of vertical angle 2 a are described. Find the 

volume of the surface thus constructed. . , . 

As. 4a6 



3 tan a 



Given a right cylinder, altitude h, and radius of base a. 
Through a diameter of the upper base two planes are 
passed, touching the lower base on opposite sides. Find 
the volume included between the planes. 

Ans. (it ]a 2 h. 



3 

6. Two cylinders of equal altitude li have a circle of radius a, 
for their common upper base. Their lower bases are 
tangent to each other. Find the volume common to 
the two cylinders. . 27 

Ans. ■ . 



CHAPTER IX. 

SUCCESSIVE INTEGRATION. 

53. Double Integral. If we reverse the operations repre- 

d 2 u 

sented by ■ , we have what is called a double integral. 

dxdy 

d 2 u 

For example, suppose = a?y 3 , 

dxdy 

then u= I I x-y 3 dydx, 

which indicates two successive integrations, the first with 
reference to x, regarding y as a constant, and the second 
with reference to y, regarding x as a constant. Thus 

omitting the constants of integration. 

54. Definite Double Integral. Here the integrations are 
between given limits. 

Tor example, 

(a-x)y 2 dydx = J (ax- — )y 2 dy 
b Jo Jb V 2 /o 



-x 



2 6 



X26 /^a 
I (a — x)y 2 dydx, the right integral sign 

with the limits (Tand a, is to be used with the variable x, and 
the left with the limits b and 2 b, with the variable y ; that is, 
the integral signs with their limits are to be taken in the same 
order as the differentials dy, dx, at the end, and from right to 



SUCCESSIVE INTEGRATION. 259 



55. Sometimes the limits of the first integration are func- 



tions of the variable of the second. 
For example, 



= 11 a 4 

24 ' 
As another example, 



Jo Jo " 2 ^ 2 ( x + 2/)^^=J r^ + fr) a2 ^ da; 

= j (a; Vet 2 — x 2 + a ~ x \dx = 



2 -x 2 \ 1 2 a? 



56. Triple Integrals. A similar notation is used for three 
successive integrations. Thus 

I I I x 2 y 2 zdxdydz = I J x 2 y 2 dxdy 

-¥Xl*-f(!"-I)-f<--^- 



EXAMPLES. 

Evaluate the following definite integrals : 

r a c h a 2 b 2 

1. « I xy(x-y)dxdy = —-(a-b). 

*/0 c/0 

2. f° fVsin OdrdO = ^^ (cos (3 - cos a), 

2/2 

•A Jo 24 



260 INTEGRAL CALCULUS. 

C* ra(l + cos9) 4 a 3 

Jo Jq o 

nm , 
^st-tfdtds = 6b\ 

s*2a s* x /*x ■ 21a 6 

7 - J. il xyzdxdydz= ~w 

8 -ij;x -*-***- v-T + * 



CHAPTER X. 

DOUBLE INTEGRATION APPLIED TO PLANE AREAS 
AND MOMENT OF INERTIA. 

57. Moment of Inertia. As an illustration of double inte- 
gration, we shall consider the problem of finding the moment 
of inertia of a given plane area. 

Definition. The moment of inertia of a given plane area 
about a given point in the plane, is the sum of the products 
obtained, by multiplying the area of each infinitesimal portion 
by the square of its distance from the given point. 



58. Double Integration. Rectangular Co-ordinates. To find 
the moment of inertia of the rectangle OACB about 0. 



Let 



0A = a, 
0B = b. 



M' N' 



rQ— 



M 



Suppose the rec- 
tangle divided into 
rectangular elements 
by lines parallel to 
the co-ordinate axes. 
Let x, y, which are to 
be regarded as independent variables, be the co-ordinates of 
any point of intersection as P, and x + dx, y + dy, the co-ordi- 
nates of Q. Then the area of the element PQ is dxdy. 

Moment of PQ = OP 2 - dxdy = (x 2 + y 2 )dxdy. 

The moment of the entire rectangle OACB is the sum of all 
the terms obtained from (x 2 -\- y 2 ) dx dy, by varying x from to 
a-, and y from to b. 

If we suppose x to be constant, while y varies from to b, we 
shall have the terms that constitute a vertical strip MNN'M'. 



262 



INTEGBAL CALCULUS. 



Hence 



Moment of MJSTJST'M' = dx C {x 2 + y 2 ) dy 



= dx (tfy + |T= (bx 2 + I) dx. 

Having thus found the moment of a vertical strip, we may 
sum all these strips, by supposing x in this result to vary from 
to a. That is, 

Moment of OACB = Cfbx* + -) dx = a ' b + ab ^ 

But the preceding operations are the same as those repre- 
sented by the double integral, 

*(x* + y 2 )dxdy. (See Art. 54.) 



ni 



If we first collect all the elements in a horizontal strip, and 
then sum these horizontal strips, we have 

Moment of OACB = C C (x 2 + y 2 ) dy dx = °* h + ah * . 
Jo «yo 3 




59. To find the moment 
of inertia of the right tri- 
angle OAG about 0. 

Let OA = a, AC=b. 

The equation of OC is 

b 
y = -x. 
u a 

This differs from the preceding problem only in the limits 
of the first integration. In collecting the elements in a vertical 
strip MN, y varies from to MN. But MN is no longer a 
constant as in Art. 58, but varies with OM, according to the 

equation of OC, y = -x. Hence the limits of y are and -x. 

In collecting all the vertical strips by the second integration, 
x varies from to a, as in Art. 58. 



APPLICATION OF DOUBLE INTEGRATION. 263 

bx 



4 + 12 



(x 2 -\-y 2 )dxdy = ab 

By supposing the triangle composed of horizontal strips as 
HK, we shall find 

Moment of 0A0 

*/0 */ay 




- ">(?+! 



60. Plane Area as a Double Integral. If in Art. 58 we omit 
the factor (x 2 + y 2 ), we shall have instead of the moment, the 
area, of the given surface. 

That is, Area = J j dx dy = I i dydx, 

the limits being determined as before. 



EXAMPLES. 

1. Find the moment of inertia about the origin, of the right 

triangle formed by the co-ordinate axes and the line 
joining the points (a, 0), (0, b). 

Ans. C C^ 1 (rf + f)dxdy = <*(<* + *) . 
Jo Jo 12 

2. Find the moment of inertia about the origin, of the circle 



x> -\-y 2 = a 2 . 



Ans. 4 



Jo Jo 



V / «2_ a ;2 



{x 2 -\-y 2 )dxdy = 



3. Find also the area of the preceding circle by Art. 60. 

Ans. 7I-0 2 . 

4. Find by Art. 60, the area between a straight line and a 

parabola, each of which joins the origin and the point 
(a, 6), the axis of X being the axis of the parabola. 

6 l/ x "•" 

Ans. f ( a dxdy= I | & dydx = ~ 

Jo Jbx Jo Jayi 6 



264 INTEGRAL CALCULUS. 

5. Find the moment of inertia of the preceding area about 
the origin, An ^ abfa? + V 



4 \1 5, 

6. Find the moment of inertia about the origin, of the area 
included within the parabola y 2 =£ax, the line x+y=3 a, 
and the axis of X. 

Ans. I |" (x 2 + y 2 ) dx cly + I I (x 2 + y 2 ) dx dy 

r 2a r Sa ~y, 2 . 9W . 314 a 4 



61. Double Integration. Polar Co-ordinates. To find the 
area of the quadrant of a circle AOB, whose radius is a. 

In rectangular co-ordinates, 
Art. 58, the lines of division 
consist of two systems, for one 
of which x is constant, and for 
the other, y is constant. 

So in polar co-ordinates, we 
have one system of straight lines 
through the pole, for each of 
which 6 is constant, and another 
system of circles about the pole as centre, for each of which r 
is constant. 

Let r, 0, which are to be regarded as independent variables, 
be the co-ordinates of any point of intersection as P, and 
r + or, + dO, the co-ordinates of Q. Then the area of PQ is 
PR* RQ = rd6.dr. 

If we first integrate regarding constant, while r varies 
from to a, we collect all the elements in any sector MOM'. 
The second integration sums all the sectors, by varying 

from to -• 

2 

Hence Area B OA = ( ( rdOdi 




Jo Jo 



APPLICATION OF DOUBLE INTEGRATION. 



265 



If we reverse the order of integration, integrating first with 
reference to 0, and afterwards with reference to r, we collect 
all the elements in a circular strip N'LL'JSf', and sum all these 
strips. This is written 



Area B OA 



Jo Jo 



r dr dO. 



62. If the moment of inertia about is required, we have 
for the moment of PQ, r 2 -rdOdr. Hence, 



Moment of BOA= C fVd0dr = C CVdrd$ 

Jo Jo Jo Jo 



ttQ? 



63. To find by a double integration the area of the semi- 
circle OB A with radius OC = a, 
the pole being on the circumfer- 
ence. 

The polar equation of the cir- 
cle is r = 2 a cos 6. If we inte- 
grate first with reference to r, then 
with reference to 6, we shall have 




Area OB A 



Jo Jo 



2a cos 



rdBdr — 



7r<r 



Here, in collecting the elements in a radial strip OM, r 
varies from to OM. But OM varies with 6, according to the 
equation of the circle r = 2acosQ. Hence the limits are and 
2 a cos 6. 

In collecting all these radial 
strips for the second integration, 

$ varies from to -• 

2 

By supposing the area composed 
of concentric circular strips about 
as LK, we find 

Area OBA= T C™ * 

Jo Jo 2 




%) 



266 INTEGRAL CALCULUS. 



EXAMPLES. 



1. Find the moment of inertia about of the area of the 

semicircle in Art. 63. A 37ra 4 

Ans. 

4 

2. Find the moment of inertia about the pole, of the area 

included by the parabola r = a sec 2 -, the initial line OX, 
and a line at right angles to it through the pole. 

Ans. Cf VcZ0dr = — • 
Jo Jo 35 

3. Find the moment of inertia about its centre, of the area 

of one loop of the lemniscate r 2 = a 2 cos 26. . ^-a/* 

nS ' 16* 

4. Find by double integration the entire area of the cardioid 

r=a(l-cos6>). An§ 3 na 2 

2 

5. Find the moment of inertia about the pole, of the area of 

the preceding cardioid. , 35 irc& 

ns . __. 



CHAPTER XL 

SURFACE AND VOLUME OP ANY SOLID. 

64. To find the area of any surface, whose equation is given 
between three rectangular co-ordinates, x, y, z. 

Let this equation be 

2=/0, 2/)- 

Suppose the given surface to be divided into elements by 
two series of planes, parallel respectively to XZ and YZ. 
These planes will also divide the plane XY into elementary 




rectangles, one of which is PQ, the projection upon the plane 
XT of the corresponding element of the surface P'Q'. 

Let x, y, z, be the co-ordinates of P\ and x + dx, y-\~dy, 
z -f dz, of Q'. 



268 INTEGRAL CALCULUS. 

Since PQ is the projection of P'Q', the area of PQ is equal 
to that of P'Q', multiplied by the cosine of the inclination of 
P'Q' to the plane XY This angle is evidently that made by 
the tangent plane at P' with the plane XY. Denoting this 
angle by y, 

Area PQ = Area P'Q' • cos y, 
Area P'Q' = Area PQ • sec y. 
We see from the figure that 

Area PQ = dx dy. 
Also from analytical geometry of three dimensions, 



»'=NS)* + (!) 



(See p. 293.) 



where — and — are partial differential coefficients, taken from 
dx dy 

the equation of the given surface z = f(x, y). 

Hence Area P'Q' = [~1 + f— Y+ (^f^dxdy. 

If S denote the required surface, 

the limits of the integration depending upon the projection, on 
the plane XY, of the surface required. 

65. For example, suppose the surface ABC to be one-eighth 
of the surface of a sphere whose equation is 

x 2 + y 2 + z 2 = a 2 . 

Here £ = -* fi = -2. 

ox z dy z 

'+(£)* + (!)*= 1+ S + H' 



SUB FACE OF ANY SOLID. 269 

' Substituting in (1) Art. 64, we have 

S = f f^dxdy = af f dxd y ■ 
J J z J J -yja 2 — x 2 — y 2 

Integrating first with reference to y, we collect all the ele- 
ments in a strip M'N'KL, y varying from zero to ML, that is, 
between the limits and Vet 2 — x 2 . 

Integrating afterwards with reference to x, we sum all the 
strips, to obtain the required surface ABC, x varying from 
to a. 

Hence 8 = a f P^ ^ - = «L 

Jo Jo Va 2 -a 2 -2/ 2 2 

EXAMPLES. 

1. The axes of two equal right circular cylinders, a being the 

radius of base, intersect at right angles ; find the sur- 
face of one intercepted by the other. 
Take for the equations of the cylinders, 

x 2 + z 2 = a 2 , and x 2 + y 2 = a 2 . 

Ans. 8«rr 8 ' =3, -^ = = 8a«. 

2. The centre of a sphere, whose radius is a, is on the sur- 

face of a right circular cylinder, the radius of whose 

base is — Find the surface of the sphere intercepted 

by the cylinder. 
Take for the equations of the sphere and cylinder, 

x 2 -\- y 2 -f z 2 = a 2 , and x 2 -\- y 2 = ax. 

Ans. 4a f (**=* dxd V - = 2Qr-2)«'. 

J° Jo Va 2 -a^-2/ 2 

3. In the preceding example, find the surface of the cylinder 

intercepted by the sphere. 

Ans. 2 a I I ■ — ==== = 4 a 2 . 

Jo Jo ^/ ax _ %2 



270 INTEGRAL CALCULUS, 

4. Find the area of that part of the surface 

2 2 -f(#cosa +2/sina) 2 = a 2 , 

which is situated in the positive compartment of co- 
ordinates. 

The surface is a right circular cylinder, whose axis is the 
line 3 = 0, iccosa + 2/sina = 0, 

and radius of base a. * a 2 



sin a cos a 



5. A diameter of a sphere, whose radius is a, is the axis of a 

right prism with a square base, 2 b being the side of the 

square. Find the surface of the sphere intercepted by 

the prism. , . & &2 

Ans. 8 a 2 6 sin x - 



Va 2 ^ 2 " a?-V 



66. To ymd' ffte volume of any solid bounded by a surface, 
whose equation is given between three rectangular co-ordinates, 
x, y, z. 

As a plane area, by dividing it into elementary rectangles, 
is 

A= J I dxdy, 

so any solid may be supposed to be divided, by planes parallel 
to the co-ordinate planes, into elementary rectangular parallelo- 
pipeds. The volume of one of these parallelopipeds is dxdydz, 
and the volume of the entire solid is 



V= ))) dxdydz, 



the limits of the integration depending upon the equation 
of the bounding surface. 



VOLUME OF ANY SOLID. 



271 



67. For example, let us find the volume of one-eighth of 
the ellipsoid, whose equation is 




PQ represents one of the elementary parallelepipeds whose 
volume is dxdydz. 

If we integrate with reference to z, we collect all the 
elements in the column MN', z varying from zero to MM 1 ; 



that is, from to z 



=4-5- 



Integrating next with reference to y, we collect all the 
columns in the slice KLN'JI, y varying from zero to KL\ 



that is, from to y = b x \l -• 

\ a 2 

This value of y is taken from the equation of the curve ALB, 






272 INTEGBAL CALCULUS. 

Finally, we integrate with, reference to x, to collect all the 
slices in the entire solid ABC. Here x varies from zero to 
OA ; that is, from to a. 

Hence we have 

V= I ( v a2 I v - h2 dxdydz. 

Jo Jo Jo 

Evaluating this integral, we find 

irabc 



V= 



6 

For the entire ellipsoid, 

-r T _ 4 7r abc 
3 

EXAMPLES. 

1. Find the volume of one of the wedges cut from the cylin- 
der, x?+ y 2 = a 2 , by the planes 

z = and z = a; tan a. 

2 a 3 tan a 



/•a /V« 2 — x 2 /*a;tana 

Ans. 2 J I dx dy dz = 



3 

2. Find the volume of the solid contained between the 
paraboloid of revolution 

x 2 -\-y 2 =az, 
the cylinder x 2 -\-y 2 =2 ax, 

and the plane 2 = 0. 

'2a /-v^Z^i /»-^- , , , 37T(X 3 



x»2a /V2ax — x2 /» — — 

^.ns. 2|| I a dxdydz 

Jo Jo Jo 



3. Find the volume bounded by the surface 

and by the positive sides of the three co-ordinate planes. 



A abc 
AnS - go' 



VOLUME OF ANY SOLID. 273 

4. The centre of a sphere of radius a, is on the surface of a 

right circular cylinder, the radius of whose base is 
-. Find the volume of the part of the cylinder inter- 
cepted by the sphere. (See Ex. 2, Art. 65.) 

5. Find the entire volume bounded by the surface, whose 

equation is x 3 -\- y% + z 3 = a*. * i^L 3 . 

0' 35 ' 






/ 






£ 



\3z 



CHAPTER XII. 

HYPERBOLIC FUNCTIONS. EQUATIONS AND PROPER^ 
TIES OF CYCLOID, EPICYCLOID, AND HYPOCYCLOID. 
INTRINSIC EQUATION OF A CURVE. 

68. We have reserved for this chapter certain miscellaneous 
subjects, for the treatment of which, both the Differential, and 
Integral, Calculus are required. 

HYPERBOLIC FUNCTIONS. 

69. Definitions. By analogy with the exponential values of 
the sine and cosine, on page 60, 

sinaj = , cosz = X . . . (1) 



the real functions 



eZ - e ", and *±* 



are called the hyperbolic sine, and hyperbolic cosine, of a?, and 
written 

e x _ g-X gX _|_ g-« 

smhic = , cosnx = — ! -. 

2 ' 2 

By substituting «V- 1 for x in (1), we find 

sinh# = ^ ~ — L cosha? = cos(a;V— 1). 

It is evident also that 

sinh = 0, cosh = 1, 

sinh (—#)= — sinh x, cosh ( — x) = cosh x. 



HYPEBBOLIC FUNCTIONS. 275 

The functions, sinh a?, cosh a;, for real values of x, are not 
periodic functions like sin#, cos a?, but increase with, x to 
infinity. 

The other hyperbolic functions are 





sinna? 


e~ — e ~ 




cosh a? 


' e x + e~*' 


ooth T — 


1 


e x -f e~ x 




tanhcc 


e x — e - ** 




1 

cosh a? 


2 




e x + e~*' 


cosecho; 


1 


2 



sinh a; e x — e~ x 

70. From these definitions we find 

cosh 2 a; — sinh 2 a; = 1, 

tanh 2 a; + sech 2 ic = 1, 

coth 2 a? — cosech 2 a; = 1, 

sinh 2 x = 2 sinh x cosh x, 

cosh 2 x — cosh 2 x + sinh 2 a?, 

sinh (a? ±y) — sinh x cosh y ± cosh a? sinh y, 

cosh (a; ± y) = cosh a? cosh 2/ ± sinh a? sinh y, 

tenh( g ±y)= tanha;±tailh y. 
1 ± tanha^tanh?/ 

71. Inverse Hyperbolic Functions. 

If a? = sinh?/, « . (1) 

then ?/ = sinh -1 x. 

But from (1), 



276 INTEGRAL CALCULUS. 

Solving this with, reference to y, 



y = log(x + V^ + l). 



Hence sinh _1 a; = log(x+ vV + 1). 

Similarly, cosh -1 x = log (x + Var 2 — 1) . 

tanli- 1 a; = -log-i^, 

2 1 — x 

coth- 1 ^ = tanh- 1 - = llog^+i 
x 2 6 x-l 



sech- 1 ^ = cosh- 1 - = log 1+Vl ~ x2 ? 

X X 



cosech -1 x = sinh -1 - = log— — ^_ 



72. Differentiation of Hyperbolic Functions. From the defi« 
nitions we have 

— sinhcc = cosh a;, 
dx 

— coshcc = sinha?, 
dx ' 

■ — tanh a? = sech 2 #, 

dx ' 

■ — cotha? = — cosech 2 a;, 
dx 

• — secha? = — sechcctanh#, 
dx 

■ — cosecha; = — cosechzcotha;. 
dx 

To differentiate the inverse function 
y = sinh -1 a;> 
we have x = sinhy, 



dx 



— = cosh v = Vsinh 2 ?/ + 1 = Vx 2 -+- 1, 
dy 



HYPERBOLIC FUNCTIONS. 



277 



dy = 1 . 

dx Va; 2 + 1 

Hence — sinh -1 x = — ■ 

dx y^ + 1 

Similarly, — cosh -1 x = — > 

dx vV-1 

Atanh.- 1 ^ 
die 



1-a; 2 

dx 1 — ar 

- — sech -1 a;= y 

dx x Vl - ar* 

— cosech -1 a; = « 

<&> £c Vl 4- a 2 



73. Inverse, Circular, and Inverse Hyperbolic, Functions as 
Integrals. A comparison of the integrals involving the inverse 
circular functions, with those involving the inverse hyperbolic 
functions, shows the close analogy between them. 



/ dx 



= sin' 



x 2 
or 






COS" 



/; 



die 



a 2 -j- a; 2 



1. _iiC 

= -tan -, 



or s-icot- 1 -. 



a 



a 



/ 



cfo 



SB 



aVsc 2 



1 ! 

- = -sec -1 -. 



/ 
/ 



Va^ + a 2 a 



da; 



or = 



cosec 






Va; 2 — 
dx 



^cosh- 1 ^. 
2 a 



-^Itanh- 1 -, 

-ar a a 

or =lcoth -1 -. 



dx 



xVd 2 — x 2 



sech -1 -- 



dx 



xVa 2 -h x 2 



— cosech -1 -. 



278 



INTEGRAL CALCULUS. 



14c. Circular and Hyperbolic Functions, as related to the Circle 
and Equilateral Hyperbola. To show the origin of the term 

hyperbolic functions, let us first 
consider the circle 

a? + y 2 = a 2 . 

If we let 

6 = angle POA, 

and u = sectorial area POA, 

we have 

a 2 
x = acosO, y = asm&, u = —— 




OM — x = a cos— -, 

a 2 

PM= y = a sin~. 
a 2 



a) 



We shall now show that if "cos" and "sin" in (1) 

are replaced by " cosh " and 
" sinh," then (1) will apply 
to the equilateral hyper- 
bola 

x*-y 2 =a\ . . (2) 

Here the sectorial area 
POA is 

a 2 , x -f- y 
V= 2 l ° S a 
(See Ex. 3, p. 246.) 

(3) 




Whence 

From (2) and (3), 



TPE 


'BB 


OiJ 


C FUNCTIONS. 


x __ 


2 m 


+ e 


2m 

^-cosh 2M 


a 




2 


a 2 


y_ 

a 


2 m 
g«* 


— e 

2 


2m 

— = smh— . 

a 2 


OM 




a? = 


,2m 
acosn— -, 
a 2 



279 



Hence 

and 
Hence 



2u 
and PM — y = a sinh — -, 

a 2 

which are similar expressions to (1). 

If 6 =■ angle POA, in the hyperbola, 

, ' V . ,2u 

tan 6 = - = tanh— - : 



hence u = — tanh" 1 tan ; 

2 ' 



whereas in the circle, 

u= 2 



-0 = -tan-Han 0. 



75. Exercises in Hyperbolic Functions. 

1 . uanh -1 • = 2 tanh -1 x. 

1-f £C 2 

2. sinh" 1 (3 x -f 4 a; 3 ) = 3 sinh" 1 x. 

3 . tanh -1 sin x = seen -1 cos x. 

4. tan -1 sinh x = sec -1 cosh #. 

5 . 2 tan -1 tanh # = tan -1 sinh 2 sc. 

6. 2 tann -1 tan a; = tanh -1 sin 2 a;. 

7. 2 cosh -1 cos x = cosh -1 cos 2 #. 

8. 2 cos -1 cosh x = cos -1 cosh 2 #. 



280 INTEGRAL CALCULUS. 

9. y = tair 1 x + tenh- 1 x. ^/ = _^ 

y dx 1-x* 

10. y = tan _i tanha\ -^ = sech2a;. 

* da; 

11. v = sinh -1 tana\ — = secsc. 

* da; 

12. y = sin~ 1 Vsin2ic — sinh.~ 1 Vsm2cc. -^ = V2cot#. 

da; 

13. w = tan _1 Vtanha; + tanh -1 Vtanha;. -^ = Vcotha;. 

da; 

14. sinha; = 3? + ^- + — + .... 

[3 [5 

15. cosha; = l-h— + —"+.... 

[2 |4 

16. tanh- 1 a;=a; + ^- + v + — • 

6 5 

as a 

17. Express the equation of the catenary ?/ = -(e a -f-e a ), 

Li 

and also the length of the arc from the vertex, in 

hyperbolic functions. 

Ans. y = acosh- y and s = asinh — 
a; a 



EQUATION AND PROPERTIES OF THE CYCLOID. 

76. Definition. The cycloid is the curve described by a 
point in the circumference of a circle, as it rolls along a straight 
line. 

Let OX be the straight line. As the circle NPT, with 
radius a, rolls along this line, the point P describes the cycloid 
OBO\ 



CYCLOID. 



281 



Let the angle through which the circle has rolled from 0, 
PCJSf=6', 
and let x, y } be the co-ordinates of P. 
Y 



, /V ,-- ; 




a W*L 



o*u 



MW * ^ 



As P is supposed to have been in contact at 0, it follows 
that 

OJST= arc PN=aO. 
Then 

x= OM= ON'-MN=a$-asmO, 

y = PM=CN 



or 



mjs = av — a sin 0, "> 
acos0 = a— acos0. ) 

If we eliminate between these equations, we have 
x = a cos -1 "v2ay — y, 

x = a vers -1 -\/2ay — y 2 






(i) 

: lL - 



(2) 



This is the equation of the cycloid, but equations (1) are 
generally more useful than (2). 

77. The point B is called the vertex of the curve. If the 
origin is transferred from to B with parallel axes, we have, 
x', y\ being the new co-ordinates, 

y — y' + 2 a, x = x' -f- ira. 



282 INTEGRAL CALCULUS. 

Substituting these in (1), we obtain 

x' = a(0 — 7r) — asin0, 

y' = — a — acos0. 

Letting — tc = 0', the angle through which the circle has 
rolled from A, and omitting the accents on x' and y', we have 



= a0' + asin0', | 
= — a + a cos 0', ) 



x = aO' + asmO', 

y 

the equation of the cycloid referred to its vertex. 



78. Tangent and Normal. From (1) Art. 76, we have 

^=a(l-cos0) = 2asin 2 ^ (1) 

dO 2 

-^ = a sin 6 = 2 a sin - cos - ; 
d<9 2 2' 

therefore ^= tan<£ = cot^ (2) 

dx 2 

Hence + = £ — -. 

A 

But since PTN=-, the angle made by PT with the axis 

6 
of X is * ; hence PT is the tangent to the curve, and PN 

Z Z 

the normal. 



79. Radius of Curvature. From (1) and (2) of the preced- 
ing article, we find 

cosec 2 - 

d?y__ zOldl^ 2 _ 1 

dy? - cosec 2 - 2 ^ 4asin2 4asill ^' 

2 2 



CYCLOID. 



283 



Substituting in the expression for the radius of curvature, p* . ' 
we have 



--(' 






1 + cot 2 - 2 4asin 4 ^ = - 4asin^ = - 2PN. 






Hence if we produce PN to Q, making NQ = PN, Q will 
be the centre of curvature for the point P. 

80. Evolute. Produce the diameter TN, making NR — TN, 
and on NR as diameter describe the circle NR. This circle 
will pass through Q, since NQ = PN. 



B 



pY 


n^s 




A 




o^y^^~ 




HQ 




^—^6 



The arc NQ = arc PN= ON, 

and arc NQR = A j 

therefore arc QR = OA - 0N= RK. 

Hence Q is a point in an equal cycloid, generated by rolling 
the circle NQR from iT along the straight line KR. 

Hence the evolute of the cycloid 0B0 1 is composed of the 
two semi-cycloids OK and IW. 



284 INTEGRAL CALCULUS. 

81. Length of Arc. To find the length of the arc OP (Fig. 
of Art. 76) we substitute in 



■JI' 



-if" 

dx 1 



dx, 



j 



-2- = cot -, and dx = 2 a sm 2 -^. 
da> 2' 2 



We thus obtain 
s 






= 2a Csm^dO = 4a/l - cos^Y 

If = 2tt, we have for the entire arc, 0-BO' = 8 a. 

This result is also evident from the property of the evolute, 

from which 

OQK=BK=4,a. 

82. Area. To find the area between the curve and the axis 
of X, we substitute in 

A=j ydx, 

y=a(l — cos6), dx = a (1 — cos 6) dQ. 
Thus we have for the entire area OBO'A, 

A = a 2 C" (1 - cos 0) 2 d0 = 3 -rra 2 . 

Hence this area is three times that of the generating circle. 

EPICYCLOID AND HYPOCYCLOID. 

83. Equation of Epicycloid. The epicycloid is the curve 
described by a point in the circumference of a circle, which 
rolls outside of a fixed circle. 

Suppose the circle BPS rolls on the fixed circle ADA', the 
point P describing the epicycloid APA'. 

Let OB = a, BC=b, BOA=cj>, BCP=^. 



EPICYCLOID AND HYPOCYCLOID. 

Since the arcs BA and BP are equal, we have 
a<f> = bif/. 



285 




*-(s-* 



O A N M 

x=OM=ON+BP 

= (a + b) cos </> -f b sin 
= (a + 5)cos<£ — &cos(^ + <£). . 
y=PM=CN-CR 

= (a + &)sin<£ — 6cos i// — (| — <£j 



= (a + 5)sin<£ — &sin(^ + <£), 
Substituting in (1) and (2), ^ = ^£, 

a; = (a + &)cos.<£ — 6 cos— ^ — <£, 
o 

y = (a + 6)sm</> — ft sin ' <£. 



(1) 



(2) 



(3) 



84. Equation of Hypocycloid. The hypocycloid is the curve 
described by a point in the circumference of a circle, which 
rolls inside of a fixed circle. 



286 



INTEGRAL CALCULUS. 



If in equations (3) Art. 83, we change b into — b, we have 
the equations of the hypocycloid, 



x = (a — b) cos <f> + b cos <j> 3 



y = (a — 6)sin<£ — 6 sin- <f>. 



a) 



85. When, in the epicycloid or hypocycloid, the ratio between 
a and b is given, we can eliminate <j> between the two equations, 
and obtain a single algebraic equation between x and y. 

For example, consider the hypocycloid where a = 4 b. Then 
equations (1) Art. 84, become 

£c = —^cos<jf> + -cos3<£ = acos 3 <£, 
y = —8in<f> sm3<£ = asm 3 <£. 



Whence X s + y 3 = (V 



as given on page 96. 



86. Radius of Curvature of Epicycloid. By differentiating 
(3) Art. 83, we have 

g = (a + 6)(sin^<£-sin<^ (1) 

.... (2) 



= 2(a + b) sin-g-4>cos a + 2 V 
v y 26^ 26 y 

-Jl = (a + 6) [ — cos "; <£ -f cos <f> 



o/ i t.\ ■ a , • a + 26 , 
2(a + 5)sm— ^sm ^ 0. 



Therefore 



cte 



26 



*. 



(3) 



EPICYCLOID AND HYPOCYCLOID. 287 

Whence 

seer — ! <f> 

d 2 y = a + 2b ^ a + 2b , d<fr = a + 26 2b 

dx> 2b ' 26 ^'cto 4&(a + 6) sin _a 6 

26^ 

Substituting in the formula for the radius of curvature, we find ' 

26 V 46(q + 5) « 

"- ^.-±26— -^+26- Mn 25* 
2b * 

a + 26 2?> a + 26 2 v ' 

If a = oo, the epicycloid becomes the cycloid, and 
a-\- b 



a + 2b 



1. 



Hence p = 45sin^, as in Art. 79. 

Li 

87. Radius of Curvature of Hypocycloid. By changing 6 
into — b in (5) Art. 86, we have for the radius of curvature 
of the hypocycloid, numerically, 

4&(a— b) . xb 

p = — * - sm~« 

H a-2b 2 

88. Length of Arc. From (2) and (4), Art. 86, we have 

Hence for the entire loop APA' (Fig. Art. 83), we have 

o/ . ja C a • a j ^j 8(a + 6)6 
s=2(a + &)| sin — <j>dd> = —± — ! — '— 

Jo 2 b a 

For the hypocycloid, the length of one loop is 
S(a-b)b 



288 INTEGBAL CALCULUS. 

89. Area betweeen Curve and Fixed Circle. To find the area 
APA'BA (Fig. Art. 83), it is better to use polar co-ordinates, 
r, 0. The formula, 

A = ±fr*d6, 

will give the area APA'OA, and this, less the area of the sec- 
tor A'OA, will be the required area. 

y 

Differentiating - = tan 0, 

x 

we have xdy -ydx = gec2 Q d$ 

x 2 

xdy — ydx — x 2 sec 2 0d0 = i^dO. 
From (3) Art. 83, and (1), (3), Art. 86, we find 



xdy — ydx = (a + b){a-{-2b)[ 1 — cos-<£ )d<j>. 



Therefore 



Cr 2 d0 = (a + b)(a + 2 b) Cfl - cos - <£ W 
Hence 



= nb(a + b)(a + 2b). 
a 

Subtracting the area of the sector 

AOA' = irab, 
we have 

~(a + b)(a + 2b)-a' 
Area APA'BA 



=4< 



7r6 2 (3a + 26) 



a 

The corresponding area for the hypocycloid is 

7rb 2 (3a-2b) 
a 



INTRINSIC EQUATION OF A CURVE. 289 

INTRINSIC EQUATION OF A CURVE. 

90. Definition. In considering the subject of curvature in 
Art. Ill, page 120, the linear motion of a point along a curve 
is compared with the corresponding change of direction. 

An equation expressing the relation between these quantities 
is called the intrinsic equation of the curve. It may be more 
precisely defined as follows : 

The intrinsic equation of a curve is the relation between the 
length of the arc measured from some fixed point, and the 
angle by which its tangent deviates from the original direction 
at the fixed point. 

It is called the intrinsic equation, because it is independent 
of any co-ordinate axes, or any external points or lines of 
reference. 

Suppose O to be the 
point of the curve from 
which the arc is meas- 
ured, and let OT be the 
tangent at 0. Taking P 
as any point of the curve, 
and letting 

s = arc OP, 

and <£ = PMT, 

the intrinsic equation will be of the form 

* =/(<£). 

The intrinsic equation of the circle whose radius is a is 
evidently s = a<j>. 

91. To find the intrinsic equation of a curve whose equation 
is given in rectangular or polar co-ordinates, it is only necessary 
to find the general expressions for s and <£, and eliminate the 
other variables. 




290 INTEGRAL CALCULUS. 

For example, let us find the intrinsic equation of the cycloid. 

Taking the vertex as origin, we use equations (1) Art. 77, 
reversing the direction of the axis of Y. We then have, 
omitting accents, 

x= a(0 + sin0), 

y = a(l — cos0). 
Differentiating these equations, we obtain 

, , dv sin# , 

tan d> = -^ = = tan — 

dx 1-I-COS0 2 

Hence <£ = ^ (1) 



^ (lJ=(U + (l)= a2(2+2cos ^ 4(l2cos 1- 

Hence s = 2ai cos-d$ = 4 a sin-. ... (2) 

•/o Z A 

Eliminating between (1) and (2), we have 

s = 4 a sin <£, 

which is the intrinsic equation of the cycloid, referred to its 
vertex. 

92. Intrinsic Equation of the Evolute. 

If we differentiate the intrinsic equation of the curve 

we have, by (1) Art. 114, Dif. Cal., the radius of curvature, 

"=!r /w (1) 

Let ! , P, be the centres of curvature for 0, P, respectively, 
and 0'P\ the evolute of OP. 

Let s=OP, <j>=PMT, 

and s'=0'P', 4>'=P'M'T'. 



INTBINSIC EQUATION OF TEE EVOLUTE. 
Since tangents to O'P' are normals to OP, 

Also s'=0'P'=PP'-00'. 

f 



291 




But from (1) PP'=/'(», 

consequently OO'=/'(0). 

Hence 8'=f'(<l>)-f'(0)= /'(<{>')- f>(0). 

Omitting the accents on s and <f>, as no longer necessary, 
we have, for the intrinsic equation of the evolute, 

93. For example, from the intrinsic equation of the cycloid 
s = 4asin<£ =/(<£), 
we have /'(<£) = 4 a cos <£, 

and /'(0) = 4a. 

Hence the equation of the evolute is 
s = 4a(cos<£ — 1), 
s being negative, as the radius of curvature is decreasing. 



292 INTEGRAL CALCULUS. 



EXAM PLES. 

Find the intrinsic equations of the following curves, and 
of their evolutes. 

X X 

1. y = ^ (e a + e °) . Ans. s = a tan <f>, and s = a tan 2 $. 

2. x 3 -\-y*=a?. Ans. s = -^sm 2 <f>, and s = -^sin2<^>. 

3. r==a(l — cos0). ^.ws. s = 4avers^, and s = ~sin2. 

3 3 3 



APPENDIX. 



ANGLES MADE WITH THE CO-ORDINATE PLANES BY THE 
TANGENT PLANE OF A SURFACE. 

94. The expression for sec y on page 268 may be derived as 
follows : 




Let P, a point of the given surface, be the point of contact. 

Through P draw PX', PT', PZ', parallel to the co-ordinate 
axes. 

Let the curve PQ be the section of the surface by the plane 
X'Z', and PM ' tangent to it ; also PR the section by the plane 
Y } Z\ and PW tangent to it. 

293 



294 INTEGRAL CALCULUS. 

Since y is constant for points in the plane X'Z', it is evi- 
dent that the tangent of the angle 31'PX' is the partial 
differential coefficient of z with respect to x ; that is, 

tan Jf'PX'= — • 
dx 

Similarly, tan NPY' = — ■ 

dy 

As the tangent plane at P contains the two tangent lines 
PM' and PN', the plane MPN' is the tangent plane. 

Pass a plane parallel to X'Y' at the distance h above it, 
intersecting the tangent lines in the points M' , N', whose 
projections are M, N. 

Draw MN, and PT perpendicular to it, and erect the plane 
PTT perpendicular to X' Y. 

Then TPT= y , 

the angle made by the tangent plane M'PJV 1 with XT'. 

Let PM= a, PN= b. 

By similar triangles 



PT:a = b: MN = b : Va 2 + b\ 



Va 2 + b 2 



tanr'Pr=^-^ Wa2 + &2 . 
PT a& 

tan 2 T'PT = ^ + ^ = tan 2 M'PM+ tan 2 JVPiT, 
a 2 6 2 



-*- 1+ (I)' + (IJ- 



TANGENT PLANE. 295 

95. Another Method. 

Let a, fi, y, be the angles made by the normal to the 
surface at P with PX', PT', PZ'. 

Let angles M'PM=A, N'PN=B. 

The direction cosines of PM' are cos A, 0, sin A ; 
ofPi^', 0, cos^ ? sin B. 

Since the normal is perpendicular to both PM' and PN\ we 
must have cos a cos A + cos y sin A = 0, 

and cos /? cos 5 + cos y sin .B = 0, 

from which cos a = — tan A cos y, 

cos /? = — tan 5 cos y. 

Substituting these expressions in 

cos 2 a + cos 2 (3 + cos 2 y = 1, 
we have cos 2 y (tan 2 J. + tan 2 B + 1) = 1, 

sec 2 y = 1 + tan 2 ^ + tan 2 £ = 1 + (— Y + f— Y- 



sec 2 


'fl : 


sec 2 y 
~~ tan 2 J. " 


sec 2 y 


sec 5 


^ 


sec 2 y 
"tan 2 B' 


sec 2 y 

"(SJ 



ADVERTISEMENTS. 



ANALYTIC GEOMETRY 

PLANE AND SOLID. 
BY E. W. NICHOLS, 

Professor of Mathematics in the Virginia Military Institute. 



The aim of the author has been to prepare a work for be- 
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For the methods of development of the various principles he has 
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preparation of the work, all authors, home and foreign, whose 
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In the first few chapters elementary examples follow the dis- 
cussion of each principle. In the subsequent chapters, sets of 
examples appear at intervals throughout each chapter, and are 
so arranged as to partake both of the nature of a review and an 
extension of the preceding principles. At the end of each 
chapter general examples, involving a more extended application 
of the principles deduced, are placed for the benefit of those 
who may desire a higher course in the subject. 

Nichols's Analytic Geometry is in use as the regular text in 
the greater number of the larger colleges and universities, and 
has proved itself adapted to the needs of institutions with the 
most varied requirements. 



Cloth. Pages xii + 275. Introduction price, $1.25. 



D. C. HEATH & CO., Publishers, Boston, New York, Chicago 



NUMBER and ITS ALGEBRA 

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In the form of a syllabus of lectures on the theory of number 
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THE NUMBER SYSTEM 
OF ALGEBRA 

Treated Theoretically and Historically 
By HENRY B. FINE, Ph.D. 

Professor of Mathematics in Princeton University 

The theoretical part of this book is an elementary exposition 
of the nature of the number concept, of the positive integer, and 
of the four artificial forms of number, which, with the positive in- 
teger, constitute the " number system " of algebra, viz., the nega- 
tive, the fraction, the irrational, and the imaginary. 

The historical part presents a resume of the history of the 
most important parts of elementary arithmetic and algebra. 

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THEORY OF EQUATIONS 

By SAMUEL MARX BARTON, Ph.D., 

Professor of Mathematics in the University of the South. 



In this treatise the author aims to give the elements of Deter- 
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special forms of determinants, followed by a collection of care- 
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upon complex numbers, properties of polynomials, general 
properties of equations, relations between roots and coefficients, 
symmetric functions, transformation of equations, limits of the 
roots of an equation, separation of roots, elimination, solution of 
numerical equations. Almost every theorem is elucidated by 
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COLLEGE ALGEBRA 

By EDWARD A. BOWSER, LL.D. 

Professor of Mathematics and Engineering in Rutgers College. 



This work is designed for academies, colleges and scientific 
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more difficult parts of the subject. 

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generally formulated in plain rules. 

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5. Systematic arrangement of material under each subject. 

6. Full notes of explanation, direction, and information, use- 
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COLLEGE ALGEBRA 

By WEBSTER WELLS, S.B., 

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of Technology. 



The first eighteen chapters have been arranged with reference 
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Attention is invited to the following particulars on account of 
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The proofs of the five fundamental laws of Algebra — the Com- 
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the Distributive Law for Multiplication — for positive or negative 
integers, and positive or negative fractions; the proofs of the 
fundamental laws of Algebra for irrational numbers ; the proof of 
the Binomial Theorem for positive integral exponents and for 
fractional and negative exponents ; the proof of Descartes's Rule 
of Signs for Positive Roots, for incomplete as well as complete 
equations ; the Graphical Representation of Functions ; the so- 
lution of Cubic and Biquadratic Equations. 

In Appendix I will be found graphical demonstrations of the 
fundamental laws of Algebra for pure imaginary and complex 
numbers ; and in Appendix II, Cauchy's proof that every equa- 
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TREATISE ON TRIGONOMETRY 

AND ITS APPLICATIONS TO 

ASTRONOMY AND GEODESY 

BY EDWARD A. BOWSER, LL.D. 

Professor of Mathematics and Engineering in Rutgers College 



The aim of the author has been to present in as concise a 
form as is consistent with clearness, the fullest course in Trigo- 
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advanced courses in colleges. 

The examples are very numerous and are carefully selected. 
Among these are some of the most elegant theorems in Plane and 
Spherical Trigonometry. The numerical solution of triangles 
has received much attention, each case being treated in detail. 

The chapters on De Moivre's Theorem, and Astronomy, 
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American text-books. 

American Mathematical Monthly : Excepting one, this is the most complete 
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to that work. In the method of treatment, arrangement, typographical execution, 
and numerous and well-selected exercises, it has no superior. The definitions of the 
functions are given " once for all " and need not be restated and modified when ob- 
tuse and reflex angles are considered. 

In the development of the theoretical part of the subject, the work is especially 
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and well-chosen collection of problems are suited to every requirement, and by 
solving these the student learns to do by doing. 

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These observations have been gathered by using the book in the class-room. 
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SURVEYING AND NAVIGATION 

AN ELEMENTARY TREATISE 

By Arthur G. Robbins, S.B., 

Assistant Professor of Civil Engineering, Massachusetts Institute of Technology 

This brief work on Surveying and Navigation is intended for 
those students who desire to supplement the study of Trigo- 
nometry with a brief course in its applications to those subjects. 
Although the subjects are treated in an elementary way, the 
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PLANE AND SPHERICAL 
TRIGONOMETRY 

BY e. miller, a.m. 

Professor of ^Mathematics and Astronomy in the University of Kansas. 



The work is arranged in a clear and logical order. It brings 
the principles of the subject face to face with operations, and 
thus not only satisfies the student of the mutual dependence 
of the two, but tends to carry him back to a clear apprehension 
of what he had probably failed to appreciate in the subordinate 
sciences. It has been remarked by many who have used the work 
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of Spherical Trigonometry ever published. 

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DESCRIPTIVE GEOMETRY 

By CLARENCE A. WALDO, Ph.D. 

Professor of Mathematics in Purdue University 



The special features of this work are : The method of develop- 
ing the subject by problems systematically arranged, and supple- 
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problems given ; the method of stating the problems, which, in 
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drawing entirely self-explanatory; the introduction of several 
subjects of considerable descriptive value, such as the axis of 
affinity, axonometry, Pascal's and Brianchon's hexagons ; the 
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GEOMETRICAL TREATMENT 
OF CURVES 

Which are isogonal conjugate to a straight line with respect to a triangle 

BY ISAAC J. SCHWATT, PH.D. 

Assistant Professor of Mathematics in the University of Pennsylvania 

The discussion includes the hyperbola and several aspects of 
the ellipse. Three large folding plates illustrate the application 
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CONIC SECTIONS 

BY rufus b. howland, b.ce. 

Professor of Mathematics in Wyoming Seminary, Kingston, Pa. 

This manual presents the elements of Conic Sections in 
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